Final Sample Question Solutions

FUN-1. Computer arithmetic

Bitwise operators and computer arithmetic can represent vectors of bits, which in turn are useful for representing sets. For example, say we have a function bit that maps elements to distinct bits; thus, bit(X) == (1 << u) for some u. Then a set {X0, X1, X2, …, Xn} can be represented as bit(X0) | bit(X1) | bit(X2) | … | bit(Xn). Element Xi is in the set with representation n if and only if (bit(Xi) & n) != 0.

QUESTION FUN-1A. What is the maximum number of set elements that can be represented in a single unsigned variable on an x86 machine?

QUESTION FUN-1B. Match each set operation with the C operator(s) that could implement that operation. (Complement is a unary operation.)

QUESTION FUN-1C. Complete this function, which should return the set difference between the sets with representations a and b. This is the set containing exactly those elements of set a that are not in set b.

unsigned set_difference(unsigned a, unsigned b) {

     return a & ~b;

OR   return a - (a & b);

OR   return a & ~(a & b);

QUESTION FUN-1D. Below we’ve given a number of C expressions, some of their values, and some of their set representations for a set of elements. For example, the first row says that the integer value of expression 0 is just 0, which corresponds to an empty set. Fill in the blanks. This will require figuring out which bits correspond to the set elements A, B, C, and D, and the values for the 32-bit int variables a, x, and s. No arithmetic operation overflows; abs(x) returns the absolute value of x (that is, x < 0 ? -x : x).

FUN-2. Bit Tac Toe

Brenda Bitdiddle is implementing tic-tac-toe using bitwise arithmetic. Her implementation starts like this:

typedef struct {
    unsigned moves[2];
} tictactoe;
#define XS 0
#define OS 1

void tictactoe_init(tictactoe* b) {
    b->moves[XS] = b->moves[OS] = 0;
}

static const unsigned ttt_values[3][3] = {
    { 0x001, 0x002, 0x004 },
    { 0x010, 0x020, 0x040 },
    { 0x100, 0x200, 0x400 }
};

`     // Mark a move by player `p` at row `row` and column `col`. `
`     // Return 0 on success; return –1 if position `row,col` has already been used. `
    int tictactoe_move(tictactoe* b, int p, int row, int col) {

1.     assert(row >= 0 && row < 3 && col >= 0 && col < 3); 2.     assert(p == XS || p == OS); 3.     /* TODO: check for position reuse */ 4.     b->moves[p] |= ttt_values[row][col]; 5.     return 0;

    }

Each position on the board is assigned a distinct bit.

QUESTION FUN-2A. Brenda’s current code doesn’t check whether a move reuses a position. Write a snippet of C code that returns –1 if an attempted move is reusing a position. This snippet will replace line 3.

if ((b->moves[XS] | b->moves[OS]) & ttt_values[row][col])
    return -1;
OR if ((b->moves[XS] | b->moves[OS] | ttt_values[row][col]) == (b->moves[XS] | b->moves[OS]))
       return -1;
OR if ((b->moves[XS] + b->moves[OS]) & ttt_values[row][col])
       return -1;
OR if ((b->moves[p] ^ ttt_values[row][col]) < b->moves[p])
       return -1;

QUESTION FUN-2B. Complete the following function. You may use the following helper function:

int popcount(unsigned n) Return the number of 1 bits in n. (Stands for “population count”; is implemented on recent x86 processors by a single instruction, popcnt.)

For full credit, your code should consist of a single “return” statement with a simple expression, but for substantial partial credit write any correct solution.

// Return the number of moves that have happened so far.
int tictactoe_nmoves(const tictactoe* b) {

    return popcount(b->moves[XS] | b->moves[OS]);

}

QUESTION FUN-2C. Write a simple expression that, if nonzero, indicates that player XS has a win on board b across the main diagonal (has marks in positions 0,0, 1,1, and 2,2).

(b->moves[XS] & 0x421) == 0x421

Lydia Davis notices Brenda’s code and has a brainstorm. “If you use different values,” she suggests, “it becomes easy to detect any win.” She suggests:

static const unsigned ttt_values[3][3] = {
    { 0x01001001, 0x00010002, 0x10100004 },
    { 0x00002010, 0x22020020, 0x00200040 },
    { 0x40004100, 0x00040200, 0x04400400 }
};

QUESTION FUN-2D. Repeat Question 1A for Lydia’s values: Write a snippet of C code that returns –1 if an attempted move is reusing a position. This snippet will replace line 3 in Brenda’s code.

if ((b->moves[XS] | b->moves[OS]) & ttt_values[row][col])
    return -1;

QUESTION FUN-2E. Repeat Question 1B for Lydia’s values: Use popcount to complete tictactoe_nmoves.

int tictactoe_nmoves(const tictactoe* b) {

    return popcount((b->moves[0] | b->moves[1]) & 0x777);

    `**`OR`**` return popcount((b->moves[0] | b->moves[1]) & 0x777000);

}

QUESTION FUN-2F. Complete the following function for Lydia’s values. For full credit, your code should consist of a single “return” statement containing exactly two constants, but for substantial partial credit write any correct solution.

` // Return nonzero if player `p` has won, 0 if `p` has not won. `
int tictactoe_check_win(const tictactoe* b, int p) {
    assert(p == XS || p == OS);

    return (b->moves[p] + 0x11111111) & 0x88888888;

// Here’s another amazing possibility (Allen Chen and others):

    return b->moves[p] & (b->moves[p] << 1) & (b->moves[p] << 2);

}

FUN-3. Data Representation

Write the value of the variable or expression in each problem -- use signed decimal representation.

For example, if we gave you:

you would write A) 10 B) -1

QUESTION FUN-3A. int i = 0xFFFF; (You may write this either in decimal or as an expression using a power of 2)

QUESTION FUN-3B. short s = 0xFFFF; (You may write this either in decimal or as an expression using a power of 2)

QUESTION FUN-3C. unsigned u = 1 \<\< 10;

QUESTION FUN-3D. From WeensyOS: unsigned long l = PTE_P \| PTE_U;

QUESTION FUN-3E. int j = ~0;

QUESTION FUN-3F. From WeensyOS: sizeof(x86_pagetable);

QUESTION FUN-3G. Given this structure:

struct s {
    char c;
    short s;
    long l;
};
struct s *ps;

This expression: sizeof(ps);

QUESTION FUN-3H. Using the structure above: sizeof(\*ps);

QUESTION FUN-3I. unsigned char u = 0xABC;

QUESTION FUN-3J. signed char c = 0xABC;

FUN-4. Memory and Pointers

Two processes are mapping a file into their address space. The mapped file contains an unsorted linked list of integers. As the processes cannot ensure that the file will be mapped at the same virtual address, they use relative pointers to link elements in the list. A relative pointer holds not an address, but an offset that user code can use to calculate a true address. Our processes define the offset as relative to the start of the file.

Thus, each element in the linked list is represented by the following structure:

struct ll_node {
    int value;
    size_t offset;
};

offset == (size_t) -1 indicates the end of the list. Other offset values represent the position of the next item in the list, calculated relative to the start of the file.

QUESTION FUN-4A. Write a function to find an item in the list. The function's prototype is:

struct ll_node* find_element(void* mapped_file, struct ll_node* list, int value);

The mapped_file parameter is the address of the mapped file data; the list parameter is a pointer to the first node in the list; and the value parameter is the value for which we are searching. The function should return a pointer to the linked list element if the value appears in the list or NULL if the value is not in the list.

struct ll_node* find_element(void* mapped_file, struct ll_node* list, int value) {
    while (1) {
        if (list->value == value)
            return list;
        if (list->offset == (size_t) -1)
            return NULL;
        list = (struct ll_node*) ((char*) mapped_file + list->offset);
    }
}

ASM-1. Data structure assembly

Here are four assembly functions, f1 through f4.

f1:
    movl    4(%esp), %eax
    movl    8(%esp), %ecx
    testl   %ecx, %ecx
    jle .L2
    xorl    %edx, %edx
.L3:
    movl    4(%eax), %eax
    incl    %edx
    cmpl    %ecx, %edx
    jne .L3
.L2:
    movl    (%edx), %eax
    ret

f2:
    movl    8(%esp), %edx
    leal    0(,%edx,4), %ecx
    movl    4(%esp), %eax
    movl    (%eax,%ecx), %eax
    addl    %ecx, %eax
    movl    (%eax), %eax
    ret

f3:
    pushl   %esi
    pushl   %ebx
    movl    12(%esp), %ecx
    movl    16(%esp), %esi
    movl    20(%esp), %eax
    testl   %esi, %esi
    jle .L9
    xorl    %edx, %edx
.L10:
    movl    %eax, %ebx
    andl    $1, %ebx
    movl    4(%ecx,%ebx,4), %ecx
    incl    %edx
    sarl    %eax
    cmpl    %esi, %edx
    jne .L10
.L9:
    movl    (%ecx), %eax
    popl    %ebx
    popl    %esi
    ret

f4:
    movl    8(%esp), %edx
    movl    4(%esp), %eax
    movl    (%eax,%edx,4), %eax
    ret

QUESTION ASM-1A. Each function returns a value loaded from some data structure. Which function uses which data structure?

1. Array
2. Array of pointers to arrays
3. Linked list
4. Binary tree

QUESTION ASM-1B. The array data structure is an array of type T. Considering the code for the function that manipulates the array, which of the following types are likely possibilities for T? Circle all that apply.

1. char
2. int
3. unsigned long
4. unsigned long long
5. char*
6. None of the above

int`, `unsigned long`, `char *

ASM-2. Where’s Waldo?

In the following questions, we give you C code and a portion of the assembly generated by some compiler for that code. (Sometimes we blank out a part of the assembly.) The C code contains a variable, constant, or function called waldo, and a point in the assembly is marked with asterisks ***. Your job is to find Waldo: write an assembly expression or constant that holds the value of waldo at the marked point. We’ve done the first one for you.

NON-QUESTION: Where’s Waldo?

int identity(int waldo) {
    return waldo;
}

00000000004007f6 `<identity>`:
  4007f6:       55                      push   %rbp
  4007f7:       48 89 e5                mov    %rsp,%rbp
  4007fa:       89 7d fc                mov    %edi,-0x4(%rbp)
  4007fd:       8b 45 fc                mov    -0x4(%rbp),%eax

           ***

  400800:       5d                      pop    %rbp
  400801:       c3                      retq   

ANSWER: %edi, -0x4(%rbp), %eax, and %rax all hold the value of waldo at the marked point, so any of them is a valid answer. If the asterisks came before the first instruction, only %edi would work.

QUESTION ASM-2A: Where’s Waldo?

int f1(int a, int b, int waldo, int d) {
    if (a > b)
        return waldo;
    else
        return d;
}

0000000000400802 `<f1>`:

  400802:       55                      push   %rbp
  400803:       48 89 e5                mov    %rsp,%rbp
  400806:       89 7d fc                mov    %edi,-0x4(%rbp)
  400809:       89 75 f8                mov    %esi,-0x8(%rbp)
  40080c:       89 55 f4                mov    %edx,-0xc(%rbp)
  40080f:       89 4d f0                mov    %ecx,-0x10(%rbp)
  400812:       8b 45 fc                mov    -0x4(%rbp),%eax
  400815:       3b 45 f8                cmp    -0x8(%rbp),%eax
  400818:       7e 05                   jle    40081f <f1+0x1d>
  40081a:       8b 45 f4                mov    -0xc(%rbp),%eax
  40081d:       eb 03                   jmp    400822 <f1+0x20>
  40081f:       8b 45 f0                mov    -0x10(%rbp),%eax
  400822:       5d                      pop    %rbp
  400823:       c3                      retq   

%edx

QUESTION ASM-2B: Where’s Waldo?

int int_array_get(int* a, int waldo) {
    int x = a[waldo];
    return x;
}

00000000004007d9 `<int_array_get>`:
INSTRUCTIONS OMITTED 

 4007dc:       8b 04 b7                mov    (%rdi,%rsi,4),%eax
 4007df:       c3                      retq   

%rsi

QUESTION ASM-2C: Where’s Waldo?

int matrix_get(int** matrix, int row, int col) {
    int* waldo = matrix[row];
    return waldo[col];
}

00000000004007e0 `<matrix_get>`:
 4007e0:       48 63 f6                movslq %esi,%rsi
 4007e3:       48 63 d2                movslq %edx,%rdx

 4007e6:       ?? ?? ?? ??             mov    ??,%rax
 4007ea:       8b 04 90                mov    (%rax,%rdx,4),%eax
 4007ed:       c3                      retq   

(%rdi,%rsi,8)

QUESTION ASM-2D: Where’s Waldo?

int f5(int x) {
    extern int waldo(int);
    return waldo(x * 45);
}

0000000000400be0 `<f5>`:

 400be0:       6b ff 2d                imul   $0x2d,%edi,%edi
 400be3:       eb eb                   jmp    400bd0

0x400bd0

QUESTION ASM-2E: Where’s Waldo?

int factorial(int waldo) {
    if (waldo < 2)
        return 1;
    else
        return waldo * factorial(waldo - 1);
}

    0000000000400910 `<factorial>`:
     400910:       83 ff 01                cmp    $0x1,%edi
     400913:       b8 01 00 00 00          mov    $0x1,%eax
     400918:       7e 13                   jle    .L2 <factorial+0x1b>
     40091a:       [6 bytes of padding (a no-op instruction)]

     400920:       0f af c7                imul   %edi,%eax
     400923:       83 ef 01                sub    $0x1,%edi
     400926:       83 ff 01                cmp    $0x1,%edi
     400929:       75 f5                   jne    .L1 <factorial+0x10>
.L2: 40092b:       f3 c3                   repz retq 

%edi

QUESTION ASM-2F: Where’s Waldo?

int binary_search(const char* needle, const char** haystack, unsigned sz) {
    unsigned waldo = 0, r = sz;
    while (waldo < r) {
        unsigned m = waldo + ((r - waldo) >> 1);
        if (strcmp(needle, haystack[m]) < 0)
            r = m;
        else if (strcmp(needle, haystack[m]) == 0)
            waldo = r = m;
        else
            waldo = m + 1;
    }
    return waldo;
}

80484ab `<binary_search>`:
     INSTRUCTIONS OMITTED
.L1: 80484c3:       89 fe                   mov    %edi,%esi
     80484c5:       29 de                   sub    %ebx,%esi
     80484c7:       d1 ee                   shr    %esi
     80484c9:       01 de                   add    %ebx,%esi
     80484cb:       8b 44 b5 00             mov    0x0(%ebp,%esi,4),%eax
     80484cf:       89 44 24 04             mov    %eax,0x4(%esp)
     80484d3:       8b 44 24 30             mov    0x30(%esp),%eax
     80484d7:       89 04 24                mov    %eax,(%esp)
     80484da:       e8 11 fe ff ff          call   80482f0 <strcmp@plt>
     80484df:       85 c0                   test   %eax,%eax
     80484e1:       78 09                   js     .L2 <binary_search+0x41>
     80484e3:       85 c0                   test   %eax,%eax
     80484e5:       74 13                   je     80484fa <binary_search+0x4f>

     80484e7:       8d 5e 01                lea    0x1(%esi),%ebx
     80484ea:       eb 02                   jmp    .L3 <binary_search+0x43>
.L2: 80484ec:       89 f7                   mov    %esi,%edi
.L3: 80484ee:       39 df                   cmp    %ebx,%edi
     80484f0:       77 d1                   ja     .L1 <binary_search+0x18>
     INSTRUCTIONS OMITTED

%ebx

In the remaining questions, you are given assembly compiled from one of the above functions by a different compiler, or at a different optimization level. Your goal is to figure out what C code corresponds to the given assembly.

QUESTION ASM-2G:

804851d `<waldo>`:
804851d:       55                      push   %ebp
804851e:       89 e5                   mov    %esp,%ebp
8048520:       83 ec 18                sub    $0x18,%esp
8048523:       83 7d 08 01             cmpl   $0x1,0x8(%ebp)
8048527:       7f 07                   jg     8048530
8048529:       b8 01 00 00 00          mov    $0x1,%eax
804852e:       eb 10                   jmp    8048540
8048530:       8b 45 08                mov    0x8(%ebp),%eax
8048533:       48                      dec    %eax
8048534:       89 04 24                mov    %eax,(%esp)
8048537:       e8 e1 ff ff ff          call   804851d
804853c:       0f af 45 08             imul   0x8(%ebp),%eax
8048540:       c9                      leave  
8048541:       c3                      ret    

What’s Waldo? Circle one.

QUESTION ASM-2H:

8048425 `<waldo>`:
8048425:       55                      push   %ebp
8048426:       89 e5                   mov    %esp,%ebp
8048428:       8b 45 08                mov    0x8(%ebp),%eax
804842b:       3b 45 0c                cmp    0xc(%ebp),%eax
804842e:       7e 05                   jle    8048435 <waldo+0x10>
8048430:       8b 45 10                mov    0x10(%ebp),%eax
8048433:       eb 03                   jmp    8048438 <waldo+0x13>
8048435:       8b 45 14                mov    0x14(%ebp),%eax
8048438:       5d                      pop    %ebp
8048439:       c3                      ret    

What’s Waldo? Circle one.

QUESTION ASM-2I:

00000000004008b4 `<waldo>`:
 4008b4:       55                      push   %rbp
 4008b5:       48 89 e5                mov    %rsp,%rbp
 4008b8:       48 83 ec 10             sub    $0x10,%rsp
 4008bc:       89 7d fc                mov    %edi,-0x4(%rbp)
 4008bf:       8b 45 fc                mov    -0x4(%rbp),%eax
 4008c2:       6b c0 2d                imul   $0x2d,%eax,%eax
 4008c5:       89 c7                   mov    %eax,%edi
 4008c7:       e8 9e 05 00 00          callq  400e6a
 4008cc:       c9                      leaveq 
 4008cd:       c3                      retq   
80484a1 `<waldo>`:
80484a1:       55                      push   %ebp
80484a2:       89 e5                   mov    %esp,%ebp
80484a4:       83 ec 18                sub    $0x18,%esp
80484a7:       8b 55 08                mov    0x8(%ebp),%edx
80484aa:       89 d0                   mov    %edx,%eax
80484ac:       c1 e0 02                shl    $0x2,%eax
80484af:       01 d0                   add    %edx,%eax
80484b1:       01 c0                   add    %eax,%eax
80484b3:       01 d0                   add    %edx,%eax
80484b5:       c1 e0 02                shl    $0x2,%eax
80484b8:       01 d0                   add    %edx,%eax
80484ba:       89 04 24                mov    %eax,(%esp)
80484bd:       e8 2b 01 00 00          call   80485ed
80484c2:       c9                      leave  
80484c3:       c3                      ret    

What’s Waldo? Circle one.

ASM-3. Disassembly I

Here’s some assembly produced by compiling a C program with gcc.

.LC1:
    .string "%d %d\n"

        .globl  f
        .type   f, @function
f:
        pushq   %rbp
        movl    $1, %ecx
.L7:
        movl    %ecx, %r8d
        movl    $1, %edx
        imull   %ecx, %r8d
.L2:
        movl    %edx, %esi
        leal    (%rdx,%rcx), %edi
        movl    $1, %eax
        imull   %edx, %esi
        addl    %r8d, %esi
.L6:
        cmpl    %edi, %eax
        jg      .L10
        movl    %eax, %r9d
        imull   %eax, %r9d
        cmpl    %r9d, %esi
        je      .L3
        incl    %eax
        jmp     .L6
.L10:
        incl    %edx
        cmpl    %edx, %ecx
        jge     .L2
        incl    %ecx
        jmp     .L7
.L3:
        pushq   %rax
        movl    $.LC0, %esi
        movl    $1, %edi
        xorl    %eax, %eax
        call    __printf_chk
        movl    $1, %eax
        popq    %rdx
        popq    %rbp
        ret

QUESTION ASM-3A. How many arguments might this function have? Circle all that apply.

1. 0
2. 1
3. 2
4. 3 or more

QUESTION ASM-3B. What might this function return? Circle all that apply.

1. 0
2. 1
3. −1
4. Its first argument, whatever that argument is
5. A square number other than 0 or 1
6. None of the above

QUESTION ASM-3C. Of these registers, which are callee-saved registers that the function saves and restores? Circle all that apply.

1. %rbx
2. %rcx
3. %rdx
4. %rbp
5. %rsi
6. %rdi
7. %r12
8. None of the above

QUESTION 3D. This function handles signed integers. If we changed the C source to use unsigned integers instead, which instructions would change? Circle all that apply.

1. movl
2. imull
3. addl
4. cmpl
5. je
6. jge
7. popl
8. None of the above

jge

QUESTION 3E. What might this function print? Circle all that apply.

1. 0 0
2. 1 1
3. 3 4
4. 4 5
5. 6 8
6. None of the above

ASM-4. Disassembly II

The questions in this section concern a function called ensmallen, which has the following assembly.

 ensmallen:

1.         movzbl  (%rsi), %edx 2.         testb   %dl, %dl 3.         movb    %dl, (%rdi) 4.         jne     .L22 5.         jmp     .L23 6. .L18: 7.         addq    $1, %rsi 8. .L22: 9.         movzbl  (%rsi), %eax 10.         cmpb    %dl, %al 11.         je      .L18 12.         addq    $1, %rdi 13.         testb   %al, %al 14.         movb    %al, (%rdi) 15.         je      .L23 16.         movl    %eax, %edx 17.         jmp     .L22 18. .L23: 19.         ret

QUESTION ASM-4A. How many arguments is this function likely to take? Give line numbers that helped you determine an answer.

QUESTION ASM-4B. Are the argument(s) pointers? Give line numbers that helped you determine an answer.

QUESTION ASM-4C. What type(s) are the argument(s) likely to have? Give line numbers that helped you determine an answer.

QUESTION ASM-4D. Write a likely signature for the function. Use return type void.

void ensmallen(unsigned char* a, unsigned char* b)

QUESTION ASM-4E. Write an alternate likely signature for the function, different from your last answer. Again, use return type void.

void ensmallen(unsigned char* a, const unsigned char* b)
void ensmallen(char* a, char* b)
void ensmallen(void* dst, const void* src)
etc., etc.

QUESTION ASM-4F. Which callee-saved registers does this function use? Give line numbers that helped you determine an answer.

QUESTION ASM-4G. The function has an “input” and an “output”. Give an “input” that would cause the CPU to jump from line 8 to label .L26, and describe what is placed in the “output” for that “input”.

QUESTION ASM-4H. Give an “input” for which the corresponding “output” is not a copy of the “input”. Your answer must differ from the previous answer.

QUESTION ASM-4I. Write C code corresponding to this function. Make it as compact as you can.

void ensmallen(char* dst, const char* src) {
    while ((*dst = *src)) {
        while (*dst == *src)
            ++src;
        ++dst;
    }
}

void ensmallen(char* dst, const char* src) {
    while (*src) {
        *dst = *src;
        while (*src == *dst)
            ++src;
        ++dst;
    }
    *dst = 0;
}

ASM-5. Machine programming

Intel really messed up this time. They’ve released a processor, the Fartium Core Trio, where every instruction is broken except the ones on this list.

(In case you forgot, xchgl swaps two values—here, the values in two registers—without modifying condition codes.)

“So what if it’s buggy,” says Andy Grove; “it can still run programs.” For instance, he argues convincingly that this function:

 void do_nothing(void) {
 }

is implemented correctly by this Fartium instruction sequence:

 ret

Your job is to implement more complex functions using only Fartium instructions. Your implementations must have the same semantics as the C functions, but may perform much worse than one might expect. You may leave off arguments and write instruction numbers (#1–12) or instruction names (for mov, use the bracketed abbreviations). Indicate where labels L1–L3 point (if you need them). Assume that on function entry, the stack is set up as on a normal x86.

QUESTION ASM-5A.

 int return_zero(void) {
     return 0;
 }

QUESTION ASM-5B.

 int identity(int a) {
     return a;
 }

QUESTION ASM-5C.

 void infinite_loop(void) {
     while (1)
         /* do nothing */;
 }

QUESTION ASM-5D.

 typedef struct point {
     int x;
     int y;
     int z;
 } point;
 
 int extract_z(point *p) {
     return p->z;
 }

 movl 4(%esp), %ecx
 xorl %eax, %eax
 incl %eax
 incl %eax
 movl (%ecx,%eax,4), %ecx
 xchgl %eax, %ecx
 ret

So much for the easy ones. Now complete one out of the following 3 questions, or more than one for extra credit. (Question 5G is worth more than the others.)

QUESTION ASM-5E.

 int add(int a, int b) {
     return a + b;
 }

     movl 4(%esp), %ecx      # %ecx := a
     movl 8(%esp), %edx      # %edx := b
     xorl %eax, %eax         # %eax := 0
     xchgl %eax, %ecx        # now %eax == a and %ecx == 0
 L3: cmpl %ecx, %edx         # compare %edx and %ecx (which is 0)
     je L1                   # "if %edx == 0 goto L1"
     incl %eax               # ++%eax
     decl %edx               # --%edx
     jmp L3
 L1: ret

 int add(int a, int b) {
     while (b != 0) {
         ++a; --b;
     }
     return a;
 }   

 int add(int a, int b) {
     int result = a;
     while (b < a) {
         ++result; ++a; --b;
     }
     return a;
 }

QUESTION ASM-5F.

 int array_dereference(int *a, int i) {
     return a[i];
 }

     movl 8(%esp), %edx        # %edx := i
     xorl %eax, %eax           # %eax := 0
 L3: xchgl %eax, %ecx
     cmpl %ecx, %edx           # compare %edx and %ecx
     xchgl %eax, %ecx
     je L1                     # "if %eax == i goto L1"
     incl %eax                 # ++%eax
     jmp L3
 L1: movl 4(%esp), %ecx        # %ecx := a
     movl (%ecx,%eax,4), %ecx  # %ecx := a[i]
     xchgl %eax, %ecx
     ret

QUESTION ASM-5G.

 int traverse_array_tree(int *a, int x) {
     int i = 0;
     while (1) {
         if (x == a[i])
             return i;
         else if (x < a[i])
             i = a[i+1];
         else
             i = a[i+2];
     }
 }

(This funky function traverses a binary tree that’s represented as an array of ints. It returns the position of the x argument in this “tree.” For example, given the following array:

 int a[] = {100, 3, 6,  50, 9, 12,  150, 0, 0,  25, 0, 0,  80, 0, 0};

the call traverse_array_tree(a, 100) returns 0, because that’s the position of 100 in a. The call traverse_array_tree(a, 80) first examines position 0; since a[0] == 100 and 80 < 100, it jumps to position a[0+1] == 3; since a[3] == 50 and 80 > 50, it jumps to position a[3+2] == 12; and then it returns 12, since a[12] == 80. The code breaks if x isn’t in the tree; don’t worry about that.)

     movl 8(%esp), %edx        # %edx := x
`      xorl %eax, %eax           # %eax := 0 (holds `i`) `
 L3: movl 4(%esp), %ecx        # %ecx := a
     movl (%ecx,%eax,4), %ecx  # %ecx := a[i]
     cmpl %ecx, %edx           # compare x and a[i]
     je L1                     # "if x == a[i] goto L1"
     jl L2                     # "if x < a[i] goto L2"
 
     incl %eax
     movl 4(%esp), %ecx
     movl (%ecx,%eax,4), %ecx
     xchgl %eax, %ecx          # i := a[i+1]
     jmp L3
 
 L2: incl %eax
     incl %eax
     movl 4(%esp), %ecx
     movl (%ecx,%eax,4), %ecx
     xchgl %eax, %ecx          # i := a[i+2]
     jmp L3
 
 L1: ret                       # return i

ASM-6. Program Layout

For the following questions, select the part(s) of memory from the list below that best describes where you will find the object.

1. heap
2. stack
3. between the heap and the stack
4. in a read-only data segment
5. in a text segment
6. in a read/write data segment
7. in a register

Assume the following code, compiled without optimization.

#include <errno.h>
#include <getopt.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>

// The following is copied from stdio.h for your reference
#define EOF (-1)

 1    unsigned long
 2    fib (unsigned long n)
 3    {
 4        if (n < 2)
 5            return (n);
 6        return (fib(n - 1) + fib(n - 2));
 7    }
 8
 9    int
10    main(int argc, char *argv[])
11    {
12        extern int optind;
13        char ch;
14        unsigned long f, n;
15
16        /* Command line processing. */
17        while ((ch = getopt(argc, argv, "h")) != EOF)
18            switch (ch) {
19            case 'h':
20            case '?':
21            default:
22                return (usage());
23            }
24
25        argc -= optind;
26        argv += optind;
27
28        if (argc != 1)
29            return (usage());
30
31        n = strtoul(strdup(argv[0]), NULL, 10);
32        if (n == 0 && errno == EINVAL)
33            return (usage());
34
35        /* Now call one of the fib routines. */
36        f = fib(n);
37        printf("fib(%lu) = %lu\n", n, f);
38
39        return (0);
40    }

QUESTION ASM-6A. The string "fib(%lu) = %lu\n" (line 37).

QUESTION ASM-6B. optind (line 25).

QUESTION ASM-6C. When executing at line 4, where you will find the address to which fib returns.

QUESTION ASM-6D. Where will you find the value of EOF that is compared to the return value of getopt in line 17.

QUESTION ASM-6E. getopt (line 17)

QUESTION ASM-6F. fib (lines 1-7)

QUESTION ASM-6G. the variable f (line 36)

QUESTION ASM-6H. the string being passed to strtoul (line 31)

QUESTION ASM-6I. strdup (line 31)

QUESTION ASM-6J. The value of the fib function when we return from fib (line 6).

ASM-7. Assembly and Data Structures

Consider the following assembly function.

func:
        xorl    %eax, %eax
        cmpb    $0, (%rdi)
        je      .L27
.L26:
        addq    $1, %rdi
        addl    $1, %eax
        cmpb    $0, (%rdi)
        jne     .L26
.L27:
        rep ret

QUESTION ASM-7A. How many parameters does this function appear to have?

QUESTION ASM-7B. What do you suppose the type of that parameter is?

QUESTION ASM-7C. Write C code that corresponds to it.

int strlen(const char* x) {
    int n = 0;
    for (; *x; ++x)
        ++n;
    return n;
}

KERN-1. Virtual memory

QUESTION KERN-1A. What is the x86 page size? Circle all that apply.

1. 4096 bytes
2. 64 cache lines
3. 512 words
4. 0x1000 bytes
5. 2¹⁶ bits
6. None of the above

The following questions concern the sizes of page tables. Answer the questions in units of pages. For instance, the page directories in 32-bit x86 WeensyOS each contained one level-1 page table page and one level-2 page table page, for a total size of 2 pages per page table.

QUESTION KERN-1B. What is the maximum size (in pages) of a 32-bit x86 page table?

QUESTION KERN-1C. What is the minimum size (in pages) of an x86 page table that would allow a process to access 2²² distinct physical addresses?

The 32-bit x86 architecture has 32-bit virtual addresses and 32-bit physical addresses. Extensions to the x86 architecture have increased both these limits.

• Physical Address Extensions (PAE) allow 32-bit machines to access up to 2⁵² bytes of physical memory (which is about 4000000 GB). That is, virtual addresses are 32 bits, and physical addresses are 52 bits.
• The x86-64 architecture evolves the x86 architecture to a 64-bit word size. x86-64 pointers are 64 bits wide instead of 32. However, only 48 of those bits are meaningful: the upper 16 bits of each virtual address are ignored. Thus, virtual addresses are 48 bits. As with PAE, physical addresses are 52 bits.

QUESTION KERN-1D. Which of these two machines would support a higher number of concurrent processes?

1. x86 with PAE with 100 GB of physical memory.
2. x86-64 with 20 GB of physical memory.

QUESTION KERN-1E. Which of these two machines would support a higher maximum number of threads per process?

1. x86 with PAE with 100 GB of physical memory.
2. x86-64 with 20 GB of physical memory.

KERN-2. Virtual memory and kernel programming

These problems consider implementations of virtual memory features in a WeensyOS-like operating system. Recall the signatures and specifications of the virtual_memory_lookup and virtual_memory_map functions:

// virtual_memory_map(pagetable, va, pa, sz, perm)
` //    Map virtual address range `[va, va+sz)` in `pagetable`. `
` //    When `X >= 0 && X < sz`, the new pagetable will map virtual address `
` //    `va+X` to physical address `pa+X` with permissions `perm`. `
//
//    Preconditions:
` //    * `va`, `pa`, and `sz` must be multiples of PAGESIZE (4096). `
//    * The level-2 pagetables referenced by the virtual address range
` //      must exist and be writable (e.g., `va + sz < MEMSIZE_VIRTUAL`). `
//
` //    Typically `perm` is a combination of `PTE_P` (the memory is Present), `
` //    `PTE_W` (the memory is Writable), and `PTE_U` (the memory may be `
` //    accessed by User applications). If `!(perm & PTE_P)`, `pa` is ignored. `
void virtual_memory_map(pageentry_t* pagetable, uintptr_t va, uintptr_t pa, size_t sz, int perm);

// virtual_memory_lookup(pagetable, va)
` //    Returns information about the mapping of the virtual address `va` in `
` //    `pagetable`. The information is returned as a `vamapping` object, `
//    which has the following components:
typedef struct vamapping {
    int pn;           // physical page number; -1 if unmapped
    uintptr_t pa;     // physical address; (uintptr_t) -1 if unmapped
    int perm;         // permissions; 0 if unmapped
} vamapping;

vamapping virtual_memory_lookup(pageentry_t* pagetable, uintptr_t va);

Also recall that WeensyOS tracks physical memory using an array of pageinfo structures:

typedef struct physical_pageinfo {
    int8_t owner;
    int8_t refcount; // 0 means the page is free
} physical_pageinfo;
static physical_pageinfo pageinfo[PAGENUMBER(MEMSIZE_PHYSICAL)];

The WeensyOS kernel occupies virtual addresses 0 through 0xFFFFF; the process address space starts at PROC_START_ADDR == 0x100000 and goes up to (but not including) MEMSIZE_VIRTUAL == 0x300000.

QUESTION KERN-2A. True or false: On x86 Linux, like on WeensyOS, the kernel occupies low virtual addresses.

QUESTION KERN-2B. On WeensyOS, which region of a process’s address space is closest to the kernel’s address space? Choose from code, data, stack, and heap.

QUESTION KERN-2C. On Linux on an x86 machine, which region of a process’s address space is closest to the kernel’s address space? Choose from code, data, stack, and heap.

Recall that the WeensyOS sys_page_alloc(addr) system call allocates a new physical page at the given virtual address. Here’s an example kernel implementation of sys_page_alloc, taken from the WeensyOS interrupt function:

case INT_SYS_PAGE_ALLOC: {
    uintptr_t addr = current->p_registers.reg_eax; // address is passed to kernel in %eax

    // [A]

    int free_pn = find_free_physical_page();
    if (free_pn < 0) { // no free physical pages
        console_printf(CPOS(24, 0), 0x0C00, "Out of physical memory!\n");
        current->p_registers.reg_eax = -1; // return result in %eax
        break; // will call run(current)
    }

    // [B]

    // otherwise, allocate the page
    assert(pageinfo[free_pn].refcount == 0);
    pageinfo[free_pn].refcount += 1;
    pageinfo[free_pn].owner = current->p_pid;

    // [C]

    // and map it into the user’s address space
    virtual_memory_map(current->p_pagetable, addr, PAGEADDRESS(free_pn), PAGESIZE, PTE_P | PTE_U | PTE_W);
    current->p_registers.reg_eax = 0;

    // [D]

    break;
}

QUESTION KERN-2D. Thanks to insufficient checking, this implementation allows a WeensyOS process to crash the operating system or even take it over. This kernel is not isolated. What the kernel should do is return −1 when the calling process supplies bad arguments. Write code that, if executed at slot [A], would preserve kernel isolation and handle bad arguments correctly.

if (addr % PAGESIZE != 0 || addr < PROC_START_ADDR || addr >= MEMSIZE_VIRTUAL) {
    current->p_registers.reg_eax = -1;
    break;
}

QUESTION KERN-2E. This implementation has another problem, which the following process would trigger:

void process_main(void) {
    heap_top = ROUNDUP((uint8_t*) end, PAGESIZE); // first address in heap region
    while (1) {
        sys_page_alloc(heap_top);
        sys_yield();
    }
}

This process code repeatedly allocates a page at the same address. What should happen is that the kernel should repeatedly deallocate the old page and replace it with a newly-allocated page. But that’s not what will happen given the example implementation.

What will happen instead? And what is the name of this kind of problem?

QUESTION KERN-2F. Write code that would fix the problem, and name the slot in the INT_SYS_PAGE_ALLOC implementation where your code should go.

vamapping vm = virtual_memory_lookup(current->p_pagetable, addr);
if (vm.perm)
    pageinfo[vm.pn].refcount -= 1;

KERN-3. Kernel programming

WeensyOS processes are quite isolated: the only way they can communicate is by using the console. Let’s design some system calls that will allow processes to explicitly share pages of memory. Then the processes can communicate by writing and reading the shared memory region. Here are two new WeensyOS system calls that allow minimal page sharing; they return 0 on success and –1 on error.

int share(pid_t p, void* addr) Allow process p to access the page at address addr.

int attach(pid_t p, void* remote_addr, void* local_addr) Access the page in process p’s address space at address remote_addr. That physical page is added to the calling process’s address space at address local_addr, replacing any page that was previously mapped there. It is an error if p has not shared the page at remote_addr with the calling process.

Here’s an initial implementation of these system calls, written as clauses in the WeensyOS kernel’s exception function.

case INT_SYS_SHARE: {
    pid_t p = current->p_registers.reg_eax;
    uintptr_t addr = current->p_registers.reg_ecx;

    // [A]

    int shindex = current->p_nshared;
    if (shindex >= MAX_NSHARED)
        goto return_error;

    // [B]

    ++current->p_nshared; 
    current->p_shared[shindex].sh_addr = addr;
    current->p_shared[shindex].sh_partner = p;
    current->p_registers.reg_eax = 0;
    break;
}

case INT_SYS_ATTACH: {
    pid_t p = current->p_registers.reg_eax;
    uintptr_t remote_addr = current->p_registers.reg_ecx;
    uintptr_t local_addr = current->p_registers.reg_edx;

    // [C]

    int shindex = -1;
    for (int i = 0; i < processes[p].p_nshared; ++i)
        if (processes[p].p_shared[i].sh_addr == remote_addr
            && processes[p].p_shared[i].sh_partner == current->p_pid)
            shindex = i;
    if (shindex == -1)
        goto return_error;

    // [D]

    vamapping vam = virtual_memory_lookup(processes[p].p_pagetable, remote_addr);

    // [E]

    virtual_memory_map(current->p_pagetable, local_addr,
                       vam.pa, PAGESIZE, PTE_P|PTE_W|PTE_U);

    // [F]

    current->p_registers.reg_eax = 0;
    break;
}

return_error:
    current->p_registers.reg_eax = -1;
    break;

Some notes:

• The implementation stores sharing records in an array. A process may call share successfully at most MAX_NSHARED times. After that, its future share calls will return an error.
• processes[p].p_nshared is initialized to 0 for all processes.
• Assume that WeensyOS has been implemented as in Problem Set 4 up through step 6 (shared read-only memory).

QUESTION KERN-3A. True or false: Given this implementation, a single WeensyOS process can cause the kernel to crash simply by calling share one or more times (with no process ever calling attach). If true, give an example of a call or calls that would likely crash the kernel.

QUESTION KERN-3B. True or false: Given this implementation, a single WeensyOS process can cause the kernel to crash simply by calling attach one or more times (with no process ever calling share). If true, give an example of a call or calls that would likely crash the kernel.

QUESTION KERN-3C. True or false: Given this implementation, WeensyOS processes 2 and 3 could work together to obtain write access to the kernel code located at address KERNEL_START_ADDR. If true, give an example of calls that would obtain this access.

#2: share(3, KERNEL_START_ADDR)
#3: attach(2, KERNEL_START_ADDR, 0x110000)

QUESTION KERN-3D. True or false: Given this implementation, WeensyOS processes 2 and 3 could work together to obtain write access to any memory, without crashing or modifying kernel code or data. If true, give an example of calls that would obtain access to a page mapped at address 0x110000 in process 5.

QUESTION KERN-3E. True or false: Given this implementation, WeensyOS child processes 2 and 3 could work together to modify the code run by a their shared parent, process 1, without crashing or modifying kernel code or data. If true, give an example of calls that would obtain write access to process 1’s code, which is mapped at address PROC_START_ADDR.

#2: share(3, PROC_START_ADDR)
#3: attach(2, PROC_START_ADDR, PROC_START_ADDR)

QUESTION KERN-3F. Every “true” answer to the preceding questions is a bug in WeensyOS’s process isolation. Fix these bugs. Write code snippets that address these problems, and say where they go in the WeensyOS code (for instance, you could refer to bracketed letters to place your snippets); or for partial credit describe what your code should do.

[A] if ((addr & 0xFFF) || addr < PROC_START_ADDR)
        return -1;

[C] if (p >= NPROC || (local_addr & 0xFFF) || local_addr < PROC_START_ADDR || local_addr >= MEMSIZE_VIRTUAL)
        return -1;

[E] if (!(vam.perm & PTE_U)
        return -1;
In virtual_memory_map: Use vam.perm instead of PTE_U|PTE_P|PTE_W

KERN-4. Teensy OS VM System

The folks at Teensy Computers, Inc, need your help with their VM system. The hardware team that developed the VM system abruptly left and the folks remaining aren't quite sure how VM works. I volunteered you to help them.

The Teensy machine has a 16-bit virtual address space with 4 KB pages. The Teensy hardware specifies a single-level page table. Each entry in the page table is 16-bits. Eight of those bits are reserved for the physical page number and 8 of the bits are reserved for flag values. Sadly, the hardware designers did not document what the bits do!

QUESTION KERN-4A. How many pages are in the Teensy virtual address space?

QUESTION KERN-4B. How many bits comprise a physical address?

QUESTION KERN-4C. Is the physical address space larger or smaller than the virtual address space?

QUESTION KERN-4D. Write, in hex, a PAGE_OFFSET_MASK (the value that when anded with an address returns the offset of the address on a page).

QUESTION KERN-4E. Write a C expression that takes a virtual address, in the variable vaddr, and returns the virtual page number.

You are now going to work with the Teensy engineers to figure out what those other bits in the page table entries mean! Fortunately, they have some engineering notes from the hardware team—they need your help in making sense of them. Each letter below has the contents of a note, state what you can conclude from that note about the lower 8 bits of the page table entries.

QUESTION KERN-4F. “Robin, I ran 8 tests using a kernel that did nothing other than loop infinitely -- for each test I set a different bit in all the PTEs of the page table. All of them ended up in the exception handler except for the one where I set bit 4. Any idea what this means?”

QUESTION KERN-4G. “Lynn, I'm writing a memory test that iterates over all of memory making sure that I can read back the same pattern I write into memory. If I don't set bit 7 of the page table entries to 1, I get permission faults. Do you know what might be happening?”

QUESTION KERN-4H. “Pat, I almost have user level processes running! It seems that the user processes take permission faults unless I have both bit 4 and bit 3 set. Do you know why?”

KERN-5. Teensy OS Page Tables

The Teensy engineers are well on their way now, but they do have a few bugs and they need your help debugging the VM system. They hand you the following page table, using the notation we used for Assignment 6 for permissions, and need your help specifying correct behavior for the operations that follow.

For each problem below, write either the physical address of the given virtual address or identify what fault would be produced. The fault types should be one of:

1. Invalid page access (there is no mapping for the requested page)
2. Privilege violation (user level process trying to access a supervisor page)
3. Permission violation (attempt to write a read-only page)

QUESTION KERN-5A. The kernel dereferences a NULL pointer

QUESTION KERN-5B. A user process dereferences a NULL pointer

QUESTION KERN-5C. The kernel writes to the address 0x8432

QUESTION KERN-5D. A user process writes to the address 0xB123

QUESTION KERN-5E. The kernel reads from the address 0x9876

QUESTION KERN-5F. A user process reads from the address 0x7654

QUESTION KERN-5G. A user process writes to the address 0xABCD

QUESTION KERN-5H. A user process writes to the address 0x2321

IO-1. Cost expressions

SH-1. Processes

This question builds versions of the existing system calls based on new abstractions. Here are three system calls that define a new abstraction called a rendezvous.

int newrendezvous(void) Returns a rendezvous ID that hasn’t been used yet.

int rendezvous(int rid, int data) Blocks the calling process P1 until some other process P2 calls rendezvous() with the same rid (rendezvous ID). Then, both of the system calls return, but P1’s system call returns P2’s data and vice versa. Thus, the two processes swap their data. Rendezvous acts pairwise; if three processes call rendezvous, then two of them will swap values and the third will block, waiting for a fourth.

void freezerendezvous(int rid, int freezedata) Freezes the rendezvous rid. All future calls to rendezvous(rid, data) will immediately return freezedata.

Here's an example. The two columns represent two processes. Assume they are the only processes using rendezvous ID 0.

This code will print

About to rendezvous
Process B got 5
Process A got 600

(the last 2 lines might appear in either order).

QUESTION SH-1A. How might you implement pipes in terms of rendezvous? Try to figure out analogues for the pipe(), close(), read(), and write() system calls (perhaps with different signatures), but only worry about reading and writing 1 character at a time.

QUESTION SH-1B. Can a rendezvous-pipe support all pipe features?

SH-2. Process management

Here’s the skeleton of a shell function implementing a simple two-command pipeline, such as “cmd1 | cmd2”.

void simple_pipe(const char* cmd1, char* const* argv1, const char* cmd2, char* const* argv2) {
    int pipefd[2], r, status;

    [A]

    pid_t child1 = fork();
    if (child1 == 0) {

        [B]

        execvp(cmd1, argv1);
    }
    assert(child1 > 0);

    [C]

    pid_t child2 = fork();
    if (child2 == 0) {

        [D]

        execvp(cmd2, argv2);
    }
    assert(child2 > 0);

    [E]

}

And here is a grab bag of system calls.

[1] close(pipefd[0]);
[2] close(pipefd[1]);
[3] dup2(pipefd[0], STDIN_FILENO);
[4] dup2(pipefd[0], STDOUT_FILENO);
[5] dup2(pipefd[1], STDIN_FILENO);
[6] dup2(pipefd[1], STDOUT_FILENO);
[7] pipe(pipefd);
[8] r = waitpid(child1, &status, 0);
[9] r = waitpid(child2, &status, 0);

Your task is to assign system call IDs, such as “1”, to slots, such as “A”, to achieve several behaviors, including a correct pipeline and several incorrect pipelines. For each question:

• You may use each system call ID once, more than once, or not at all.
• You may use zero or more system call IDs per slot. Write them in the order they should appear in the code.
• You may assume that no signals are delivered to the shell process (so no system call ever returns an EINTR error).
• The function should wait for both commands in the pipeline to complete before returning.
• It may help to detach the last “Reference material” page of the exam.

QUESTION SH-2A. Implement a correct foreground pipeline.

QUESTION SH-2B. Implement a pipeline so that, given arguments corresponding to “echo foo | wc -c”, the wc process reads “foo” from its standard input but does not exit thereafter. For partial credit describe in words how this might happen.

QUESTION SH-2C. Implement a pipeline so that, given arguments corresponding to “echo foo | wc -c”, “foo” is printed to the shell’s standard output and the wc process prints “0”. (In a correctly implemented pipeline, “wc” would print 4, which is the number of characters in “foo\n”.) For partial credit describe in words how this might happen.

QUESTION SH-2D. Implement a pipeline that appears to work correctly on “echo foo | wc -c”, but always blocks forever if the left-hand command outputs more than 65536 characters. For partial credit describe in words how this might happen.

QUESTION SH-2E. Implement a pipeline so that, given arguments corresponding to “echo foo | wc -c”, both echo and wc report a “Bad file descriptor” error. (This error, which corresponds to EBADF, is returned when a file descriptor is not valid or does not support the requested operation.) For partial credit describe in words how this might happen.

SH-3. Processes

Consider the two programs shown below.

// Program 1
#include <stdio.h>
#include <unistd.h>
int
main(void)
{
    printf("PID %d running prog1\n", getpid());
}

// Program 2
#include <stdio.h>
#include <unistd.h>
int
main(void)
{
    char *argv[2];
    argv[0] = "prog1";
    argv[1] = NULL;
    printf("PID %d running prog2\n", getpid());
    int r = execv("./prog1", argv);
    printf("PID %d exiting from prog2\n", getpid());
}

QUESTION SH-3A. How many different PIDs will print out if you run Program 2?

QUESTION SH-3B. How many lines of output will you see?

Now, let's assume that we change Program 2 to the following:

// Program 2B
#include <stdio.h>
#include <unistd.h>
int
main(void)
{
    char* argv[2];
    argv[0] = "prog1";
    argv[1] = NULL;

    printf("PID %d running prog2\n", getpid());
    pid_t p = fork();
    if (p == 0) {
        int r = execv("./prog1", argv);
    } else {
        printf("PID %d exiting from prog2\n", getpid());
    }
}

QUESTION SH-3C. How many different PIDs will print out if you run Program 2B?

QUESTION SH-3D. How many lines of output will you see?

Finally, consider this version of Program 2.

// Program 2C
#include <stdio.h>
#include <unistd.h>
int
main(void)
{
    char *argv[2];
    argv[0] = "prog1";
    argv[1] = NULL;

    printf("PID %d running prog2\n", getpid());
    pid_t p = fork();
    pid_t q = fork();
    if (p == 0 || q == 0) {
        int r = execv("./prog1", argv);
    } else {
        printf("PID %d exiting from prog2\n", getpid());
    }
}

QUESTION SH-3E. How many different PIDs will print out if you run Program 2C?

QUESTION SH-3F. How many lines of output will you see?

SH-4. Be a CS61 TF!

You are a CS61 teaching fellow. A student working on A4 is having difficulty getting pipes working. S/he comes to you for assistance. The function below is intended to traverse a linked list of commands, fork/exec the indicated processes, and hook up the pipes between commands correctly. The student has commented it reasonably, but is quite confused about how to finish writing the code. Can you help? Figure out what code to add at points A, B, and C.

#include "sh61.h"

typedef struct command command;
struct command {
    command *next; // Next in sequence of commands
    int argc;      // number of arguments
    int ispipe;    // pipe symbol follows this command
    char** argv;   // arguments, terminated by NULL
    pid_t pid;     // pid running this command
};

void
do_pipes(command *c)
{
    pid_t newpid;
    int havepipe = 0;   // We had a pipe on the previous command
    int lastpipe[2]= {-1, -1};
    int curpipe[2];

    do {
        if (c->ispipe)
            assert(pipe(curpipe) == 0);

        newpid = fork();
        switch (newpid) {
            case 0:
                if (havepipe) {
                    // There was a pipe on the last command; It's stored
                    // in lastpipe; I need to hook it up to this process???
                    // **** PART A ****
                }
                if (c->ispipe) {
                    // The current command is a pipe -- how do I hook it up???
                    // **** PART B ****
                }

                execvp(c->argv[0], c->argv);
                
                fprintf(stderr, "Exec failed\n");
                c->pid = -1;
                break;
            case -1:
                c->pid = newpid;
                break;
            default:
                // I bet there is some cleanup I have to do here!?
                // **** PART C ****

                // Set up for the next command
                havepipe = c->ispipe;
                if (c->ispipe) {
                    lastpipe[0] = curpipe[0];
                    lastpipe[1] = curpipe[1];
                }
                c->pid = newpid;
                c = c->next;
                break;
        }
    } while (newpid != -1 && havepipe);
}

QUESTION SH-4A. What should go in the Part A space above, in anything?

close(lastpipe[1]);
dup2(lastpipe[0], STDIN_FILENO);
close(lastpipe[0]);

QUESTION SH-4B. What should go in the Part B space above, in anything?

close(curpipe[0]);
dup2(curpipe[1], STDOUT_FILENO);
close(curpipe[1]);

QUESTION SH-4C. What should go in the Part C space above, in anything?

if (havepipe) {
    close(lastpipe[0]);
    close(lastpipe[1]);
}

NET-1. Networking

QUESTION NET-1A. Which of the following system calls should a programmer expect to sometimes block (i.e., to return after significant delay)? Circle all that apply.

QUESTION NET-1B. Below are seven message sequence diagrams demonstrating the operation of a client–server RPC protocol. A request such as “get(X)” means “fetch the value of the object named X”; the response contains that value. Match each network property or programming strategy below with the diagram with which it best corresponds. You will use every diagram once.

A	B	C	D

E	F	G

NET-2. Making Network Servers Robust

QUESTION NET-2A. You've built a network server, list the resources that you might run out of if someone launched a DoS attack on you.

QUESTION NET-2B. Sam suggests that you just create a separate thread to handle each incoming connection. Why isn't this necessarily going to work?

QUESTION NET-2C. A server sets up a socket to listen on a connection. When a client wants to establish a connection, how does the server manage the multiple clients? In your answer indicate what system call or calls are used and what they do.

QUESTION NET-2D. Which of the following system calls might block?

• accept
• bind
• connect
• listen
• setsockopt
• select
• socket

SYNCH-1. Threads

The following code performs a matrix multiplication, c = ab, where a, b, and c are all square matrices of dimension sz. It uses the cache-friendly ikj index ordering.

#define MELT(matrix, sz, row, col) (matrix)[(row)*(sz) + (col)]

void matrix_multiply(double* c, const double* a, const double* b, size_t sz) {
    for (size_t i = 0; i < sz; ++i)
        for (size_t j = 0; j < sz; ++j)
            MELT(c, sz, i, j) = 0;
    for (size_t i = 0; i < sz; ++i)
        for (size_t k = 0; k < sz; ++k)
            for (size_t j = 0; j < sz; ++j)
                MELT(c, sz, i, j) += MELT(a, sz, i, k) * MELT(b, sz, k, j);
}

But matrix multiplication is a naturally parallelizable problem. Here’s some code that uses threads to multiply even faster on a multicore machine. We use sz parallel threads, one per row of c.

     typedef struct matrix_args {
         double* c;
         const double* a;
         const double* b;
         size_t sz;
         size_t i;
     } matrix_args;
 
     void* matrix_multiply_ikj_thread(void* arg) {
 (α)     matrix_args* m = (matrix_args*) arg;
 (β)     for (size_t j = 0; j < m->sz; ++j)
 (γ)         MELT(m->c, m->sz, m->i, j) = 0;
 (δ)     for (size_t k = 0; k < m->sz; ++k)
 (ε)         for (size_t j = 0; j < m->sz; ++j)
 (ζ)             MELT(m->c, m->sz, m->i, j) += MELT(m->a, m->sz, m->i, k) * MELT(m->b, m->sz, k, j);
 (η)     return NULL;
      }
 
     void matrix_multiply_ikj(double* c, const double* a, const double* b, size_t sz) {
 (1)     pthread_t* threads = (pthread_t*) malloc(sizeof(pthread_t) * sz);
 (2)     for (size_t i = 0; i < sz; ++i) {
 (3)         matrix_args m = { c, a, b, sz, i };
 (4)         int r = pthread_create(&threads[i], NULL, &matrix_multiply_ikj_thread, &m);
 (5)         assert(r == 0);
 (6)     }
 (7)     for (size_t i = 0; i < sz; ++i)
 (8)         pthread_join(threads[i], NULL);
 (9)     free(threads);
     }

But when run, this code gives wildly incorrect results.

QUESTION SYNCH-1A. What is wrong? Describe why the problem is a synchronization issue.

QUESTION SYNCH-1B. Write C code showing how the problem could be fixed with changes only to matrix_multiply_ikj. Refer to the numbered lines to indicate replacements and/or insertions. Use one or more additional heap allocations and no additional calls to pthread functions. Free any memory you allocate once it is safe to do so.

     void matrix_multiply_ikj(double *c, const double *a, const double *b, size_t sz) {
 (1)     pthread_t *threads = (pthread_t *) malloc(sizeof(pthread_t) * sz);
   *     matrix_args *marr = (matrix_args *) malloc(sizeof(matrix_args) * sz);         
 (2)     for (size_t i = 0; i < sz; ++i) {
 (3)         matrix_args m = { c, a, b, sz, i };
   *         marr[i] = m;
   *         int r = pthread_create(&threads[i], NULL, &matrix_multiply_ikj_thread,
   *                                &marr[i]);
 (5)         assert(r == 0);
 (6)     }
 (7)     for (size_t i = 0; i < sz; ++i)
 (8)         pthread_join(threads[i], NULL);
 (9)     free(threads);
   *     free(marr);
     }

On single-core machines, the kij order performs almost as fast as the ikj order. Here’s a version of the parallel matrix multiplication code that uses kij.

     typedef struct matrix_args_kij {
         double* c;
         const double* a;
         const double* b;
         size_t sz;
         size_t k;
     } matrix_args_kij;
 
     void* matrix_multiply_kij_thread(void* arg) {
 (α)     matrix_args_kij* m = (matrix_args_kij*) arg;
 (β)     for (size_t i = 0; i < m->sz; ++i)
 (γ)         for (size_t j = 0; j < m->sz; ++j)
 (δ)             MELT(m->c, m->sz, i, j) += MELT(m->a, m->sz, i, m->k) * MELT(m->b, m->sz, m->k, j);
 (ε)     return NULL;
      }
 
     void matrix_multiply_kij(double* c, const double* a, const double* b, size_t sz) {
 (1)     pthread_t* threads = (pthread_t*) malloc(sizeof(pthread_t) * sz);
 (2)     for (size_t i = 0; i < sz; ++i)
 (3)         for (size_t j = 0; j < sz; ++j)
 (4)             MELT(c, sz, i, j) = 0;
 (5)     for (size_t k = 0; k < sz; ++k) {
 (6)         matrix_args_kij m = { c, a, b, sz, k };
 (7)         int r = pthread_create(&threads[k], NULL, &matrix_multiply_kij_thread, &m);
 (8)         assert(r == 0);
 (9)     }
(10)     for (size_t k = 0; k < sz; ++k)
(11)         pthread_join(threads[k], NULL);
(12)     free(threads);
     }

This problem has the same problem as the previous version, plus another problem. Even after your fix from 8A–8B is applied, this version produces incorrect results.

QUESTION SYNCH-1C. What is the new problem? Describe why it is a synchronization issue.

QUESTION SYNCH-1D. Write pseudocode or C code that fixes this problem. You should refer to pthread functions. For full credit your solution should have low contention.

     typedef struct matrix_args_kij {
         double *c;
         const double *a;
         const double *b;
  *      pthread_mutex_t *locks;
         size_t sz;
         size_t k;
     } matrix_args;
 
     void *matrix_multiply_kij_thread(void *arg) {
         matrix_args_kij *m = (matrix_args_kij *) arg;
         for (size_t i = 0; i < m->sz; ++i)
             for (size_t j = 0; j < m->sz; ++j) {
  *              pthread_mutex_lock(&m->locks[i * m->sz + j]);
                 MELT(m->c, m->sz, i, j) += MELT(m->a, m->sz, i, m->k) * MELT(m->b, m->sz, m->k, j);
  *              pthread_mutex_unlock(&m->locks[i * m->sz + j]);
             }
         return NULL;
      }
 
     void matrix_multiply_kij(double *c, const double *a, const double *b, size_t sz) {
         pthread_t *threads = (pthread_t *) malloc(sizeof(pthread_t) * sz);
  *      matrix_args_kij *marr = (matrix_args_kij *) malloc(sizeof(matrix_args_kij) * sz);
  *      pthread_mutex_t *locks = (pthread_mutex_t *) malloc(sizeof(pthread_mutex_t) * sz * sz);
         for (size_t i = 0; i < sz; ++i)
             for (size_t j = 0; j < sz; ++j) {
                 MELT(c, sz, i, j) = 0;
  *              pthread_mutex_init(&locks[i * sz + j], NULL);
             }
         for (size_t k = 0; k < sz; ++k) {
  *          matrix_args_kij m = { c, a, b, locks, sz, k };
  *          marr[k] = m;
  *          int r = pthread_create(&threads[k], NULL, &matrix_multiply_kij_thread,
                                    &marr[k]);
             assert(r == 0);
         }
         for (size_t k = 0; k < sz; ++k)
             pthread_join(threads[k], NULL);
         free(threads);
  *      free(marr);
  *      free(locks);
     }

SYNCH-2. Synchronization and concurrency

Most synchronization objects have at least two operations. Mutual-exclusion locks support lock and unlock; condition variables support wait and signal; and from section notes you may remember the semaphore synchronization object, one of the earliest synchronization objects ever invented, which supports P and V.

In this problem, you’ll work with a synchronization object with only one operation, which we call a hemiphore. Hemiphores behave like the following; it is very important that you understand this pseudocode.

typedef struct hemiphore {
    int value;
} hemiphore;

// Initialize the hemiphore to value 0.
void hemiphore_init(hemiphore* h) {
    h->value = 0;
}

` // Block until the hemiphore has value >= `bound`, then  ``**`atomically`**``  increment its value by `delta`. `
void H(hemiphore* h, int bound, int delta) {
    // This is pseudocode; a real hemiphore implementation would block, not spin, and would
    // ensure that the test and the increment happen in one atomic step.
    while (h->value < bound)
        sched_yield();
    h->value += delta;
}

Once a hemiphore is initialized with hemiphore_init, application code should access the hemiphore only through the H operation.

QUESTION SYNCH-2A. Use hemiphores to implement mutual-exclusion locks. Fill out the code below. (You may not need to fill in every empty slot. You may use standard C constants; for example, INT_MIN is the smallest possible value for a variable of type int, which on a 32-bit machine is −2147483648.)

typedef struct mutex {
    hemiphore h;
} mutex;

void mutex_init(mutex* m) {
    hemiphore_init(&m->h);
}

void mutex_lock(mutex* m) {

}

void mutex_unlock(mutex* m) {

}

QUESTION SYNCH-2B. Use hemiphores to implement condition variables. Fill out the code below. You may assume that the implementation of mutex is your hemiphore-based implementation from above (so, for instance, cond_wait may access the hemiphore m->h). See the Hints at the end of the question.

typedef struct condvar {
    mutex m;
    hemiphore h;

} condvar;

void cond_init(condvar* c) {
    mutex_init(&c->m);
    hemiphore_init(&c->h);

}

void cond_signal(condvar* c) {

}

void cond_wait(condvar* c, mutex* m) {

}

typedef struct condvar {
    hemiphore h;
} condvar;

void cond_init(condvar* c) {
    hemiphore_init(&c->h);
}

void cond_signal(condvar* c) {

}

void cond_wait(condvar* c, mutex* m) {

}

Hints. For full credit:

• If no thread is waiting on condition variable c, then cond_signal(c) should do nothing.
• Assume N threads are waiting on condition variable c. Then N calls to cond_signal(c) are both necessary and sufficient to wake them all up.
• Your solution must not add new sleep-wakeup race conditions to the user’s code. (That is, no sleep-wakeup race conditions unless the user uses mutexes incorrectly.)

QUESTION SYNCH-2C. Use pthread mutexes and condition variables to implement hemiphores. Fill out the code below. See the hints after the question.

typedef struct hemiphore {
    pthread_mutex_t m;
    int value;
    pthread_cond_t c;
} hemiphore;

void hemiphore_init(hemiphore* h) {
    pthread_mutex_init(&h->m);
    h->value = 0;
    pthread_cond_init(&h->c);
}

void H(hemiphore* h, int bound, int delta) {

}

Hints. The pthread mutex and condition variable operations have the following signatures. You should pass NULL for any attributes arguments. Don’t worry about the pthread_mutex_destroy and pthread_cond_destroy operations, and feel free to abbreviate (e.g. “lock” instead of “pthread_mutex_lock”).

• pthread_mutex_init(pthread_mutex_t* m, const pthread_mutexattr_t* attributes)
• pthread_mutex_lock(pthread_mutex_t* m)
• pthread_mutex_unlock(pthread_mutex_t* m)
• pthread_cond_init(pthread_cond_t* c, const pthread_condattr_t* attributes)
• pthread_cond_signal(pthread_cond_t* c) (wakes up at most one thread waiting on c)
• pthread_cond_broadcast(pthread_cond_t* c) (wakes up all threads waiting on c)
• pthread_cond_wait(pthread_cond_t* c, pthread_mutex_t* m)

QUESTION SYNCH-2D. Consider the following two threads, which use a shared hemiphore h with initial value 0.

Thread 1                      Thread 2

H(&h, 1000, 1);               while (1) {
printf("Thread 1 done\n");        H(&h, 0, 1);
                                  H(&h, 0, -1);
                              }

Thread 2 will never block, and the hemiphore’s value will alternate between 1 and 0. Thread 1 will never reach the printf, because the hemiphore’s value never reaches 1000. However, in most people’s first implementation of hemiphores using pthread mutexes and condition variables, Thread 1 will not block. Every call to H in Thread 2 will effectively wake up Thread 1. Though Thread 1 will then check the hemiphore’s value and immediately go back to sleep, doing so wastes CPU time.

Design an implementation of hemiphores using pthread mutexes and condition variables that solves this problem. In your revised implementation, Thread 1 above should block forever. For full credit, write C code. For partial credit, write pseudocode or English describing your design.

Hint. One working implementation constructs a linked list of “waiter” objects, where each waiter object is on a different thread’s stack, as initially sketched below. You can use such objects or not as you please.

typedef struct hemiphore_waiter {
    struct hemiphore_waiter* next;

} hemiphore_waiter;

typedef struct hemiphore {
    pthread_mutex_t m;
    int value;
    hemiphore_waiter* waiters;
} hemiphore;

void hemiphore_init(hemiphore* h) {
    pthread_mutex_init(&h->m);
    h->value = 0;
    h->waiters = NULL;
}

void H(hemiphore* h, int bound, int delta) {
    hemiphore_waiter w;

    `**`pthread_cond_init(&w.c);`**`                // no points off if missing

    // no points off for linked list messups

}

typedef struct hemiphore {
    pthread_mutex_t m;
    int value;

} hemiphore;

void hemiphore_init(hemiphore* h) {
    pthread_mutex_init(&h->m);
    h->value = 0;

}

void H(hemiphore* h, int bound, int delta) {

}

SYNCH-3. Pipes and synchronization

In the following questions, you will implement a mutex using a pipe, and a limited type of pipe using a mutex.

The definitions of the pthread mutex and condition variable operations are as follows.

int pthread_mutex_init(pthread_mutex_t* m, const pthread_mutexattr_t* attr) Create a new mutex with attributes defined by attr. (For this question, attr is ignored.)

int pthread_mutex_lock(pthread_mutex_t* m) Locks m. If the mutex is already locked, the calling thread will block until the mutex becomes available.

int pthread_mutex_unlock(pthread_mutex_t* m) Unlocks m. Calling pthread_mutex_unlock with a mutex that the calling thread does not hold will result in undefined behavior.

int pthread_cond_init(pthread_cond_t* c, const pthread_condattr_t* attr) Create a new condition variable with attributes defined by attr. (For this question, attr is ignored.)

int pthread_cond_signal(pthread_cond_t* c) Unblocks one thread waiting for c.

int pthread_cond_wait(pthread_cond_t* c, pthread_mutex_t* m) Atomically unlocks m and blocks the calling thread on the condition c. When the condition is signaled, the thread locks m and returns. Calling pthread_cond_wait with an unlocked mutex will result in undefined behavior.

The operations return 0 on success. Although errors are possible (for instance, ENOMEM if there’s not enough memory to allocate a new mutex) you may assume that they don’t occur.

QUESTION SYNCH-3A. In this question, you are to implement mutex functionality using a pipe. Fill in the definitions of pipe_mutex_init, pipe_mutex_lock, and pipe_mutex_unlock. You should be able to implement the same functionality as the pthread versions (assuming no other code accesses the pipe).

typedef struct pipe_mutex {
    int fd[2];
} pipe_mutex;

int pipe_mutex_init(pipe_mutex* m) {     
    if (pipe(&m->fd) < 0)
        return -1;

    write(m->fd[1], "X", 1);

    return 0;
}

int pipe_mutex_lock(pipe_mutex* m) {        

    char buf;
    read(m->fd[0], &buf, 1);

}

int pipe_mutex_unlock(pipe_mutex* m) {       

    write(m->fd[1], "X", 1);

}

In the next questions, you will help implement pipe functionality using an in-memory buffer and a mutex. This “mutex pipe” will only work between threads of the same process (in contrast to a regular pipe, which also works between processes). An initial implementation of mutex pipes is as follows; you will note that it contains no mutexes.

    typedef struct mutex_pipe {

1.    char buf[BUFSIZ]; 2.    size_t head; 3.    size_t sz;

    } mutex_pipe;

    int mutex_pipe_init(mutex_pipe* p) {

6.    p->head = p->sz = 0; 7.    memset(&p->buf[0], 0, sizeof(p->buf)); 8.    return 0;

    }

`     // Read up to `sz` bytes from the mutex_pipe into `buf` and return the number of bytes `
    // read. If no bytes are available, wait until at least one byte can be read.
    ssize_t mutex_pipe_read(mutex_pipe* p, char* buf, size_t sz) {

10.    size_t n = 0; 11.    while (n < sz && (p->sz != 0 || n == 0)) { 12.        size_t ncopy = p->sz; 13.        if (ncopy > sizeof(p->buf) - p->head) 14.            ncopy = sizeof(p->buf) - p->head; 15.        if (ncopy > sz - n) 16.            ncopy = sz - n; 17.        memcpy(&buf[n], &p->buf[p->head], ncopy); 18.        n += ncopy; 19.        p->head += ncopy; 20.        p->head = p->head % sizeof(p->buf); 21.        p->sz -= ncopy; 22.    } 23.    return n;

    }

`     // Write up to `sz` bytes from `buf` into the mutex_pipe and return the number of bytes `
    // written. If no space is available, wait until at least one byte can be written.
    ssize_t mutex_pipe_write(mutex_pipe* p, const char* buf, size_t sz) {

30.    size_t n = 0; 31.    while (n < sz && (p->sz != sizeof(p->buf) || n == 0)) { 32.        size_t tail = p->head + p->sz; 33.        tail = tail % sizeof(p->buf); 34.        size_t ncopy = sizeof(p->buf) - p->sz; 35.        if (ncopy > sizeof(p->buf) - tail) 36.            ncopy = sizeof(p->buf) - tail; 37.        if (ncopy > sz - n) 38.            ncopy = sz - n; 39.        memcpy(&p->buf[tail], &buf[n], ncopy); 40.        n += ncopy; 41.        p->sz += ncopy; 42.    } 43.    return n;

    }

The last page of this exam has a copy of that code that you can remove and keep.

NOT A QUESTION. It would be wise to work through an example. For example, assume BUFSIZ == 4, and figure out how the following calls would behave.

mutex_pipe_write(p, "Hi", 2);
mutex_pipe_read(p, buf, 4);
mutex_pipe_write(p, "Test", 4);
mutex_pipe_read(p, buf, 3);

First let’s reason about this code in the absence of threads.

QUESTION SYNCH-3B. Which of the following changes could, if made in isolation, result in undefined behavior when a mutex pipe was used? Circle all that apply.

1. Eliminating line 6
2. Eliminating line 7
3. Eliminating lines 13–14
4. Eliminating lines 15–16
5. Eliminating line 18
6. Eliminating line 19

QUESTION SYNCH-3C. Which of the following changes could, if made in isolation, cause a mutex_pipe_read to return incorrect data (that is, the byte sequence produced by read will not equal the byte sequence passed to write)? Circle all that apply.

1. Eliminating line 33
2. Eliminating lines 35–36
3. Eliminating lines 37–38
4. Eliminating line 39
5. Eliminating line 40
6. Eliminating line 41

QUESTION SYNCH-3D. Which of the following changes could, if made in isolation, cause a call to mutex_pipe_write to never return (when a correct implementation would return)? Circle all that apply.

1. Eliminating line 33
2. Eliminating lines 35–36
3. Eliminating lines 37–38
4. Eliminating line 39
5. Eliminating line 40
6. Eliminating line 41

QUESTION SYNCH-3E. Write an invariant for p->sz. An invariant is a statement about the value of p->sz that is always true. Write your invariant in the form of an assertion; for full credit give the most specific true invariant you can. (“p->sz is an integer” is unspecific, but true; “p->sz == 4” is specific, but false.)

assert(  p->sz >= 0 && p->sz <= BUFSIZ  );

QUESTION SYNCH-3F. Write an invariant for p->head. For full credit give the most specific true invariant you can.

assert(  p->head >= 0 && p->head < BUFSIZ  );

In the remaining questions, you will add synchronization objects and operations to make your mutex pipe work in a multithreaded program. Here is your starting point:

    typedef struct mutex_pipe {

1.    char buf[BUFSIZ]; 2.    size_t head; 3.    size_t sz; 4.    pthread_mutex_t m;

    } mutex_pipe;

    int mutex_pipe_init(mutex_pipe* p) {

5.    pthread_mutex_init(&p->m, NULL); 6.    p->head = p->sz = 0; 7.    memset(&p->buf[0], 0, sizeof(p->buf)); 8.    return 0;

    }

    (the rest of the code as in the prior questions)

QUESTION SYNCH-3G. Add calls to “lock” (pthread_mutex_lock) and “unlock” (pthread_mutex_unlock) that protect the mutex pipe from race condition bugs. Write one or more snippets of C code and give line numbers after which the snippets should appear. For full credit, your solution must not deadlock—if one thread is reading from a pipe and another thread is writing to the pipe, then both threads must eventually make progress.

• Add pthread_mutex_lock(&p->m); after lines: 10, 30
• Add pthread_mutex_unlock(&p->m); after lines: 22, 42
• Other changes (if any):

if (ncopy == 0) {
    pthread_mutex_unlock(&p->m);   // or just "unlock"
    sched_yield();   // optional
    pthread_mutex_lock(&p->m);     // or just "lock"
}

if (n == 0) {
    pthread_mutex_unlock(&p->m);   // or just "unlock"
    sched_yield();   // optional
    pthread_mutex_lock(&p->m);     // or just "lock"
}

QUESTION SYNCH-3H. Your solution to the last question has poor utilization. For instance, a thread that calls mutex_pipe_read on an empty mutex pipe will spin forever, rather than block. Introduce one or more condition variables so that mutex_pipe_read will block until data is available. Write one or more snippets of C code and give line numbers after which the snippets should appear.

• Add to struct mutex_pipe: pthread_cond_t c;
• Add to mutex_pipe_init after line 7: pthread_cond_init(&c, NULL);
• Other changes:

if (ncopy == 0)
    pthread_cond_wait(&p->c, &p->m);

if (n != 0)
    pthread_cond_signal(&p->c);

SYNCH-4. Race conditions

Most operating systems support process priority levels, where the kernel runs higher-priority processes more frequently than lower-priority processes. A hypothetical Unix-like operating system called “Boonix” has two priority levels, normal and batch. A Boonix parent process changes the priority level of one of its children with this system call:

int setbatch(pid_t p) Sets process p to have batch priority. All future children of p will also have batch priority. Returns 0 on success, –1 on error. Errors include ESRCH, if p is not a child of the calling process.

Note that a process cannot change its own batch status.

You’re writing a Boonix shell that can run commands with batch priority. If c->isbatch is nonzero, then c should run with batch priority, as should its children. Your start_command function looks like this:

    pid_t start_command(command* c) {

1.    c->pid = fork(); 2.    if (c->pid == 0) { 3.        handle_pipes(c); 4.        handle_redirections(c); 5.        (void) execvp(c->argv[0], c->argv); 6.        // if we get here, execvp failed 7.        perror("execvp"); 8.        exit(1); 9.    } 10.    assert(c->pid > 0); 11.    if (c->isbatch) 12.        setbatch(c->pid); 13.    return c->pid;

    }

This shell has two race conditions, one more serious.

QUESTION SYNCH-4A. In some cases, c will change to batch priority after it starts running. Draw a dependency diagram demonstrating this race condition, or briefly describe it.

QUESTION SYNCH-4B. In some cases, c or one of its children could run forever with normal priority. Draw a dependency diagram demonstrating this race condition, or briefly describe it.

In the remaining questions, you will fix these race conditions in three different ways. The first uses a new system call:

int isbatch() Returns 1 if the calling process has batch priority, 0 if it has normal priority.

QUESTION SYNCH-4C. Use isbatch to prevent both race conditions. Write a snippet of C code and give the line number after which it should appear. You should need one code snippet.

while (c->isbatch && !isbatch())
    /* spin */;

QUESTION SYNCH-4D. Use the pipe system call and friends to prevent both race conditions. Write snippets of C code and give the line numbers after which they should appear. You should need several snippets. Make sure you clean up any extraneous file descriptors before running the command or returning from start_command.

int pipefd[2];
pipe(pipefd);

if (c->isbatch) {
    char buf;
    read(pipefd[0], &buf, 1);
}
close(pipefd[0]);
close(pipefd[1]);

write(pipefd[1], "X", 1);
close(pipefd[0]);
close(pipefd[1]);

int pipefd[2];
pipe(pipefd);

close(pipefd[1]);
if (c->isbatch)
    read(pipefd[0], &pipefd, 1);  // won’t ever read any chars, so this is safe
close(pipefd[0]);

close(pipefd[0]);
close(pipefd[1]);

QUESTION SYNCH-4E. Why should the pipe solution be preferred to the isbatch solution? A sentence, or the right single word, will suffice.

QUESTION SYNCH-4F. Suggest a change to the setbatch system call’s behavior that could fix both race conditions, and say how to use this new setbatch in start_command. Write one or more snippets of C code and give the line numbers after which they should appear.

if (c->isbatch)
    setbatch(getpid());

MISC-1. Debugging

In the following short-answer questions, you have access to five debugging tools: top, strace, gdb, valgrind, and man. You can’t change program source code or use other tools. Answer the questions briefly (a couple sentences at most).

QUESTION MISC-1A. You are given a program that appears to “get stuck” when run. How would you distinguish whether the program blocked forever (e.g., made a system call that never returned) or entered an infinite loop?

QUESTION MISC-1B. You are given a program that uses a lot of memory. How would you tell whether the program leaks memory?

QUESTION MISC-1C. You are given a program that produces weird answers. How would you check if it invoked undefined behavior?

QUESTION MISC-1D. You are given a program that blocks forever. How would you tell where the program blocked (which function called the blocking system call)?

QUESTION MISC-1E. You are given a program that takes a long time to produce a result. How would you tell whether the program was using system calls unintelligently?

QUESTION MISC-1F. You are given a program that exits with a system call error, but doesn’t explain what happened in detail. How would you find what error condition occurred and understand the conditions that could cause that error?

MISC-2. Miscellany

QUESTION MISC-2A. True or false in conventional Unix systems?

1. File descriptors are often used to communicate among processes on the same machine.

   True

2. File descriptors are often used to communicate among processes on different machines.

   True

3. File descriptors are often used to communicate with persistent storage.

   True

4. File descriptors are often used to access primary memory.

   False

5. File descriptors are often used to create child processes.

   False

QUESTION MISC-2B. Match the process isolation feature on the left with the hardware feature that helps enforce it on the right. Use each hardware feature once (make the best match you can).

The remaining questions refer to the following lines of code.

1.   close(fd); 2.   connect(fd, sockaddr, socklen); 3.   listen(fd); 4.   mmap(NULL, 4096, PROT_READ, MAP_SHARED, fd, 0); 5.   read(fd, buf, 4096); 6.   write(fd, buf, 4096);

QUESTION MISC-2C. If a program executes the following line without error, which lines could be executed next without error? List all numbers that apply.

fd = open("/home/cs61user/cs61-psets/pset6/pong61.c", O_RDWR);

QUESTION MISC-2D. If a program executes the following line without error, which lines could be executed next without error? List all numbers that apply.

fd = socket(AF_INET, SOCK_STREAM, 0);

QUESTION MISC-2E. If a program executes the following lines without error, which lines could be executed next without error? List all numbers that apply.

pipe(pipefd); fd = pipefd[0];

MISC-3. More Miscellany

QUESTION MISC-3A. True or false: Any C arithmetic operation has a well-defined result.

QUESTION MISC-3B. True or false: Any x86 processor instruction has a well-defined result.

QUESTION MISC-3C. True or false: By executing a trap instruction, a process can force an operating system kernel to execute arbitrary code.

QUESTION MISC-3D. True or false: By manipulating process memory and registers, an operating system kernel can force a process to execute arbitrary instructions.

QUESTION MISC-3E. True or false: All signals are sent explicitly via the kill() system call.

QUESTION MISC-3F. True or false: An operating system’s buffer cache is generally fully associative.

QUESTION MISC-3G. True or false: The least-recently-used eviction policy is more useful for very large files that are read sequentially than it is for stacks.

QUESTION MISC-3H. True or false: Making a cache bigger can lower its hit rate for a given workload.

QUESTION MISC-3I. True or false: x86 processor caches are coherent (i.e., always appear to contain the most up-to-date values).

QUESTION MISC-3J. True or false: A socket file descriptor supports either reading or writing, but not both.

MISC-4. Pot Pourri

Parts A-D pertain to the data structures and hexdump output shown here.

struct x {
    unsigned long ul;
    unsigned short us;
    unsigned char uc;
} *sp;

// Hexdump output of some program running on the appliance
08c1b008  e9 11 cf d0 0d d0 3f f3  63 61 74 00 0d f0 fe ca  |......?.cat.....|
08c1b018  5e ea 15 0d de c0 ad de                           |^.......|

You are told that sp = 0x08c1b008.

QUESTION MISC-4A. What is the value (in hex) of sp->ul?

QUESTION MISC-4B. What is the value (in hex) of sp->uc?

QUESTION MISC-4C. At what address will you find the string "cat"?

QUESTION MISC-4D. You think that the bytes after the string "cat" comprise an array of 3 integers; what is the value (in hex) of the middle one of those?

QUESTION MISC-4E. What is the following binary value expressed in hexadecimal: 01011010?

QUESTION MISC-4F. What is the value of the hex number 0x7FF in decimal?

QUESTION MISC-4G. Is 0x98765432 a valid return from malloc?

QUESTION MISC-4H. What is the minimum number of x86 instruction bytes you need to write an infinite loop?

QUESTION MISC-4I. True or False: Every declaration in C code allocates space for an object.

QUESTION MISC-4J. True or False: Processes cannot share memory.

For parts K–O, assume we are running on the appliance and we initialize ival, p, and q as shown below. Write the value of the expression -- you may express the values in hex if that's simpler, just be sure to prefix them with 0x to make it clear that you are doing so. For True/False questions, there is no need to correct or provide a counterexample for any statements that are false.

int ival[4] = {0x12345678, 0x9ABCDEF0, 0x13579BDF,0x2468ACE0};
int* p = &ival[0];
int* q = &ival[3];
int* x = p + 1;
char* cp = (char*) (q - 2);

QUESTION MISC-4K. q - p

QUESTION MISC-4L. ((char \*)q - (char \*)p)

QUESTION MISC-4M. x - p

QUESTION MISC-4N. \*((short \*)((char \*)x+2))

QUESTION MISC-4O. \*cp

QUESTION MISC-4P. What system call allows you to block on a collection of file descriptors?

QUESTION MISC-4Q. What system call creates a communication channel that can only be used among related processes?

pipe

QUESTION MISC-4R. What system call can change the attributes of a file descriptor so you can poll on it rather than block?

fcntl

QUESTION MISC-4S. What system call produces a file descriptor on which a server can exchange messages with a client?

socket

QUESTION MISC-4T. True or False: A program and a process are the same thing.