This is not the current version of the class.

Data representation exercises

Exercises that seem less appropriate this year, or which cover topics that we haven’t covered in class, are generally marked with ⚠️.

DATAREP-1. Sizes and alignments

QUESTION DATAREP-1A. True or false: For any non-array type X, the size of X (sizeof(X)) is greater than or equal to the alignment of type X (alignof(X)).

QUESTION DATAREP-1B. True or false: For any type X, the size of struct Y { X a; char newc; } is greater than the size of X.

QUESTION DATAREP-1C. True or false: For any types A1...An (with n ≥ 1), the size of struct Y is greater than the size of struct X, given:

struct X {
    A1 a1;
    ...
    An an;
};

struct Y {
    A1 a1;
    ...
    An an;
    char newc;
};

QUESTION DATAREP-1D. True or false: For any types A1...An (with n ≥ 1), the size of struct Y is greater than the size of union X, given:

union X {
    A1 a1;
    ...
    An an;
};

struct Y {
    A1 a1;
    ...
    An an;
};

QUESTION DATAREP-1E. Assume that structure struct Y { ... } contains K char members and M int members, with K≤M, and nothing else. Write an expression defining the maximum sizeof(struct Y).

QUESTION DATAREP-1F. You are given a structure struct Z { T1 a; T2 b; T3 c; } that contains no padding. What does (sizeof(T1) + sizeof(T2) + sizeof(T3)) % alignof(struct Z) equal?

QUESTION DATAREP-1G. Arrange the following types in increasing order by size. Sample answer: “1 < 2 = 4 < 3” (choose this if #1 has smaller size than #2, which has equal size to #4, which has smaller size than #3).

  1. char
  2. struct minipoint { uint8_t x; uint8_t y; uint8_t z; }
  3. int
  4. unsigned short[1]
  5. char**
  6. double[0]

DATAREP-2. Expressions

QUESTION DATAREP-2A. Here are eight expressions. Group the expressions into four pairs so that the two expressions in each pair have the same value, and each pair has a different value from every other pair. There is one unique answer that meets these constraints. m has the same type and value everywhere it appears (there’s one unique value for m that meets the problem’s constraints). Assume an x86-32 machine: a 32-bit architecture in which pointers are 32 bits long.

  1. sizeof(&m)
  2. -1
  3. m & -m
  4. m + ~m + 1
  5. 16 >> 2
  6. m & ~m
  7. m
  8. 1

DATAREP-3. Hello binary

This problem locates 8-bit numbers horizontally and vertically in the following 16x16 image. Black pixels represent 1 bits and white pixels represent 0 bits. For horizontal arrangements, the most significant bit is on the left as usual. For vertical arrangements, the most significant bit is on top.

A pixelated cat

Examples: The 8-bit number 15 (hexadecimal 0x0F, binary 0b00001111) is located horizontally at 3,4, which means X=3, Y=4.

15 is also located horizontally at 7,6.

The 8-bit number 0 is located vertically at 0,0. It is also located horizontally at 0,0 and 1,0.

The 8-bit number 134 (hexadecimal 0x86, binary 0b10000110) is located vertically at 8,4.

QUESTION DATAREP-3A. Where is 3 located vertically? (All questions refer to 8-bit numbers.)

QUESTION DATAREP-3B. Where is 12 located horizontally?

QUESTION DATAREP-3C. Where is 255 located vertically?

DATAREP-4. Hello memory

Shintaro Tsuji wants to represent the image of Question DATAREP-3 in computer memory. He stores it in an array of 16-bit unsigned integers:

unsigned short cute[16];

Row Y of the image is stored in integer cute[Y].

QUESTION DATAREP-4A. What is sizeof(cute), 2, 16, 32, or 64?

QUESTION DATAREP-4B. printf("%d\n", cute[0]); prints 16384. Is Shintaro’s machine big-endian or little-endian?

DATAREP-5. Hello program

Now that Shintaro has represented the image in memory as an array of unsigned short objects, he can manipulate the image using C. For example, here’s a function.

void swap(void) {
    for (int i = 0; i < 16; ++i) {
        cute[i] = (cute[i] << 8) | (cute[i] >> 8);
    }
}

Running swap produces the following image:

Swapped cat

Shintaro has written several other functions. Here are some images (A is the original):

Cat A Cat B Cat C Cat D Cat E

A

B

C

D

E
Cat F Cat G Cat H Cat I Cat J

F

G

H

I

J

For each function, what image does that function create?

QUESTION DATAREP-5A.

void f0() {
    for (int i = 0; i < 16; ++i) {
        cute[i] = ~cute[i];
    }
}

QUESTION DATAREP-5B.

void f1() {
    for (int i = 0; i < 16; ++i) {
        cute[i] = ((cute[i] >> 1) & 0x5555) | ((cute[i] << 1) & 0xAAAA);
        cute[i] = ((cute[i] >> 2) & 0x3333) | ((cute[i] << 2) & 0xCCCC);
        cute[i] = ((cute[i] >> 4) & 0x0F0F) | ((cute[i] << 4) & 0xF0F0);
        cute[i] =  (cute[i] >> 8)           |  (cute[i] << 8);
    }
}

QUESTION DATAREP-5C.

void f2() {
    char* x = (char*) cute;
    for (int i = 0; i < 16; ++i) {
        x[2*i] = i;
    }
}

For “fun”

The following programs generated the other images. Can you match them with their images?

void f3() {
    for (int i = 0; i < 16; ++i) {
        cute[i] &= ~(7 << i);
    }
}

void f4() {
    swap();
    for (int i = 0; i < 16; ++i) {
        cute[i] <<= i/4;
    }
    swap();
}

void f5() {
    for (int i = 0; i < 16; ++i) {
        cute[i] = -1 * !!(cute[i] & 64);
    }
}

void f6() {
    for (int i = 0; i < 8; ++i) {
        int tmp = cute[15-i];
        cute[15-i] = cute[i];
        cute[i] = tmp;
    }
}

void f7() {
    for (int i = 0; i < 16; ++i) {
        cute[i] = cute[i] & -cute[i];
    }
}

void f8() {
    for (int i = 0; i < 16; ++i) {
        cute[i] ^= cute[i] ^ cute[i];
    }
}

void f9() {
    for (int i = 0; i < 16; ++i) {
        cute[i] = cute[i] ^ 4080;
    }
}

DATAREP-6. Memory regions

Consider the following program:

struct ptrs {
    int** x;
    int* y;
};

struct ptrs global;

void setup(struct ptrs* p) {
    int* a = malloc(sizeof(int));
    int* b = malloc(sizeof(int));
    int* c = malloc(sizeof(int));
    int* d = malloc(sizeof(int));
    int* e = malloc(sizeof(int) * 2);
    int** f = malloc(4 * sizeof(int*));
    int** g = malloc(sizeof(int*));

    *a = 0;
    *b = 0;
    *c = (int) a;
    *d = *b;
    e[0] = 29;
    e[1] = (int) &d[100000];

    f[0] = b;
    f[1] = c;
    f[2] = 0;
    f[3] = 0;

    *g = c;

    global.x = f;
    global.y = e;

    p->x = g;
    p->y = &e[1];
}

int main(int argc, char** argv) {
    stack_bottom = (char*) &argc;
    struct ptrs p;
    setup(&p);
    m61_collect();
    do_stuff(&p);
}

This program allocates objects a through g on the heap and then stores those pointers in some stack and global variables. (It then calls our conservative garbage collector from class⚠️, but that won’t matter until the next problem.) We recommend you draw a picture of the state setup creates.

QUESTION DATAREP-6A. Assume that (uintptr_t) a == 0x8300000, and that malloc returns increasing addresses. Match each address to the most likely expression with that address value. The expressions are evaluated within the context of main. You will not reuse an expression.

Value         Expression
1. 0x8300040 A. &p
2. 0x8049894 B. (int*) *p.x[0]
3. 0x8361AF0 C. &global.y
4. 0x8300000 D. global.y
5. 0xBFAE0CD8 E. (int*) *p.y

DATAREP-7. Garbage collection

Here is the top-level function for the conservative garbage collector we wrote in class. (⚠️ 2018 note: We haven’t done this. ⚠️)

void m61_collect(void) {
    char* stack_top = (char*) &stack_top;

    // The entire contents of the heap start out unmarked
    for (size_t i = 0; i != nmr; ++i) {
        mr[i].marked = 0;
    }

    // Mark all reachable objects, starting with the roots (the stack)
    m61_markaccessible(stack_top, stack_bottom - stack_top);

    // Free everything that wasn't marked
    for (size_t i = 0; i != nmr; ++i) {
        if (mr[i].marked == 0) {
            m61_free(mr[i].ptr);
            --i;                // m61_free moved different data into this
                                // slot, so we must recheck the slot
        }
    }
}

This garbage collector is not correct because it doesn’t capture all memory roots.

Consider the program from the previous section, and assume that an object is reachable if do_stuff can access an address within the object via variable references and memory dereferences without casts or pointer arithmetic. Then:

QUESTION DATAREP-7A. Which reachable objects will m61_collect() free? Circle all that apply.

a

b

c

d

e

f

g

None of these

QUESTION DATAREP-7B. Which unreachable objects will m61_collect() not free? Circle all that apply.

a

b

c

d

e

f

g

None of these

QUESTION DATAREP-7C. Conservative garbage collection in C is often slower than precise garbage collection in languages such as Java. Why? Circle all that apply.

  1. C is generally slower than other languages.
  2. Conservative garbage collectors must search all reachable memory for pointers. Precise garbage collectors can ignore values that do not contain pointers, such as large character buffers.
  3. C programs generally use the heap more than programs in other languages.
  4. None of the above.

DATAREP-8. Memory errors

The following function constructs and returns a lower-triangular matrix of size N. The elements are random 2-dimensional points in the unit square. The matrix is represented as an array of pointers to arrays.

struct point2 {
    double d[2];
};
typedef point2* point2_vector;

point2_vector* make_random_lt_matrix(size_t N) {
    point2_vector* m = (point2_vector*) malloc(sizeof(point2_vector) * N);
    for (size_t i = 0; i < N; ++i) {
        m[i] = (point2*) malloc(sizeof(point2) * (i + 1));  /* LINE A */
        for (size_t j = 0; j <= i; ++j) {
            for (int d = 0; d < 2; ++d) {
                m[i][j].d[d] = drand48();                   /* LINE B */
            }
        }
    }
    return m;
}

This code is running on an x86-32 machine (size_t is 32 bits, not 64). You may assume that the machine has enough free physical memory and the process has enough available virtual address space to satisfy any memory allocation request.

QUESTION DATAREP-8A. Give a value of N so that, while make_random_lt_matrix(N) is running, no new fails, but a memory error (such as a null pointer dereference or an out-of-bounds dereference) happens on Line A. The memory error should happen specifically when i == 1.

(This problem is probably easier when you write your answer in hexadecimal.)

QUESTION DATAREP-8B. Give a value of N so that no new fails, and no memory error happens on Line A, but a memory error does happen on Line B.

DATAREP-9. Data representation

Assume a 64-bit x86-64 architecture unless explicitly told otherwise.

Write your assumptions if a problem seems unclear, and write down your reasoning for partial credit.

QUESTION DATAREP-9A. Arrange the following values in increasing numeric order. Assume that x is an int with value 8192.

a. EOF e. 1000
b. x & ~x f. (signed char) 65535
c. (signed char) 0x47F g. The size of the stdio cache
d. x | ~x h. (int) 0x80000000U

A possible answer might be “a < b < c = d < e < f < g < h.”

For each of the remaining questions, write one or more arguments that, when passed to the provided function, will cause it to return the integer 61 (which is 0x3d hexadecimal). Write the expected number of arguments of the expected types.

QUESTION DATAREP-9B.

int f1(int n) {
    return 0x11 ^ n;
}

QUESTION DATAREP-9C.

int f2(const char* s) {
    return strtol(s, nullptr, 0);
}

QUESTION DATAREP-9D. Your answer should be different from the previous answer.

int f3(const char* s) {
    return strtol(s, nullptr, 0);
}

QUESTION DATAREP-9E. For this problem, you will also need to define a global variable. Give its type and value.

f4:
    andl $5, %edi
    leal (%rsi,%rdi,2), %eax
    movzbl y(%rip), %ecx
    subl %ecx, %eax
    retq

DATAREP-10. Sizes and alignments

Assume a 64-bit x86-64 architecture unless explicitly told otherwise.

Write your assumptions if a problem seems unclear, and write down your reasoning for partial credit.

QUESTION DATAREP-10A. Use the following members to create a struct of size 16, using each member exactly once, and putting char a first; or say “impossible” if this is impossible.

  1. char a; (we’ve written this for you)
  2. unsigned char b;
  3. short c;
  4. int d;
struct size_16 {
    char a;




};

QUESTION DATAREP-10B. Repeat Part A, but create a struct with size 12.

struct size_12 {
    char a;




};

QUESTION DATAREP-10C. Repeat Part A, but create a struct with size 8.

struct size_8 {
    char a;




};

QUESTION DATAREP-10D. Consider the following structs:

struct x {
    T x1;
    U x2;
};
struct y {
    struct x y1;
    V y2;
};

Give definitions for T, U, and V so that there is one byte of padding in struct x after x2, and two bytes of padding in struct y after y1.

DATAREP-11. Dynamic memory allocation

QUESTION DATAREP-11A. True or false?

  1. free(nullptr) is an error.
  2. malloc(0) can never return nullptr.

QUESTION DATAREP-11B. Give values for sz and nmemb so that calloc(sz, nmemb) will always return nullptr (on a 32-bit x86 machine), but malloc(sz * nmemb) might or might not return null.

Consider the following 8 statements. (p and q have type char*.)

  1. free(p);
  2. free(q);
  3. p = q;
  4. q = nullptr;
  5. p = (char*) malloc(12);
  6. q = (char*) malloc(8);
  7. p[8] = 0;
  8. q[4] = 0;

QUESTION DATAREP-11C. Put the statements in an order that would execute without error or evoking undefined behavior. Memory leaks count as errors. Use each statement exactly once. Sample answer: “abcdefgh.”

QUESTION DATAREP-11D. Put the statements in an order that would cause one double-free error, and no other error or undefined behavior (except possibly one memory leak). Use each statement exactly once.

QUESTION DATAREP-11E. Put the statements in an order that would cause one memory leak (one allocated piece of memory is not freed), and no other error or undefined behavior. Use each statement exactly once.

QUESTION DATAREP-11F. Put the statements in an order that would cause one boundary write error, and no other error or undefined behavior. Use each statement exactly once.

DATAREP-12. Pointers and debugging allocators

You are debugging some students’ m61 code from Problem Set 1. The codes use the following metadata:

struct meta { ...
    meta* next;
    meta* prev;
};

meta* mhead;    // head of active allocations list

Their linked-list manipulations in m61_malloc are similar.

void* m61_malloc(size_t sz, const char* file, int line) {
    ...
    meta* m = (meta*) ptr;
    m->next = mhead;
    m->prev = nullptr;
    if (mhead) {
        mhead->prev = m;
    }
    mhead = m;
    ...
}

But their linked-list manipulations in m61_free differ.

Alice’s code:

void m61_free(void* ptr, ...) { ...
    meta* m = (meta*) ptr - 1;
    if (m->next != nullptr) {
        m->next->prev = m->prev;
    }
    if (m->prev == nullptr) {
        mhead = nullptr;
    } else {
        m->prev->next = m->next;
    }
    ...
}

Bob’s code:

void m61_free(void* ptr, ...) { ...
    meta* m = (meta*) ptr - 1;
    if (m->next) {
        m->next->prev = m->prev;
    }
    if (m->prev) {
        m->prev->next = m->next;
    }
    ...
}

Chris’s code:

void m61_free(void* ptr, ...) { ...
    meta* m = (meta*) ptr - 1;
    m->next->prev = m->prev;
    m->prev->next = m->next;
    ...
}

Donna’s code:

void m61_free(void* ptr, ...) { ...
    meta* m = (meta*) ptr - 1;
    if (m->next) {
        m->next->prev = m->prev;
    }
    if (m->prev) {
        m->prev->next = m->next;
    } else {
        mhead = m->next;
    }
    ...
}

You may assume that all code not shown is correct.

QUESTION DATAREP-12A. Whose code will segmentation fault on this input? List all students that apply.

int main() {
    void* ptr = malloc(1);
    free(ptr);
}

QUESTION DATAREP-12B. Whose code might report something like “invalid free of pointer [ptr1], not allocated” on this input? (Because a list traversal starting from mhead fails to find ptr1.) List all students that apply. Don’t include students whose code would segfault before the report.

int main() {
    void* ptr1 = malloc(1);
    void* ptr2 = malloc(1);
    free(ptr2);
    free(ptr1);   // <- message printed here
}

QUESTION DATAREP-12C. Whose code would improperly report something like “LEAK CHECK: allocated object [ptr1] with size 1” on this input? (Because the mhead list appears not empty, although it should be.) List all students that apply. Don’t include students whose code would segfault before the report.

int main() {
    void* ptr1 = malloc(1);
    free(ptr1);
    m61_printleakreport();
}

QUESTION DATAREP-12D. Whose linked-list code is correct for all inputs? List all that apply.

DATAREP-13. Arena allocation

Chimamanda Ngozi Adichie is a writing a program that needs to allocate and free a lot of nodes, where a node is defined as follows:

struct node {
    int key;
    void* value;
    node* left;
    node* right;      // also used in free list
};

She uses an arena allocator variant. Here’s her code.

struct arena_group {
    arena_group* next_group;
    node nodes[1024];
};

struct arena {
    node* frees;
    arena_group* groups;
};

node* node_alloc(arena* a) {
    if (!a->frees) {
        arena_group* g = new arena_group;
        // ... link `g` to `a->groups` ...
        for (size_t i = 0; i != 1023; ++i) {
            g->nodes[i].right = &g->nodes[i + 1];
        }
        g->nodes[1023].right = nullptr;
        a->frees = &g->nodes[0];
    }
    node* n = a->frees;
    a->frees = n->right;
    return n;
}

void node_free(arena* a, node* n) {
    n->right = a->frees;
    a->frees = n;
}

QUESTION DATAREP-13A. True or false?

  1. This allocator never has external fragmentation.
  2. This allocator never has internal fragmentation.

QUESTION DATAREP-13B. Chimamanda’s frenemy Paul Auster notices that if many nodes are allocated right in a row, every 1024th allocation seems much more expensive than the others. The reason is that every 1024th allocation initializes a new group, which in turn adds 1024 nodes to the free list. Chimamanda decides instead to allow a single element of the free list to represent many contiguous free nodes. The average allocation might get a tiny bit slower, but no allocation will be much slower than average. Here’s the start of her idea:

node* node_alloc(arena* a) {
    if (!a->frees) {
        arena_group* g = new arena_group;
        // ... link `g` to `a->groups` ...
        g->nodes[0].key = 1024;   // g->nodes[0] is the 1st of 1024 contiguous free nodes
        g->nodes[0].right = nullptr;
        a->frees = &g->nodes[0];
    }
    node* n = a->frees;
    // ???
    return n;
}

Complete this function by writing code to replace // ???.

QUESTION DATAREP-13C. Write a node_free function that works with the node_alloc function from the previous question.

void node_free(arena* a, node* n) {




}

QUESTION DATAREP-13D. Complete the following new function.

// Return the arena_group containing node `n`. `n` must be a node returned by
// a previous call to `node_alloc(a)`.
arena_group* node_find_group(arena* a, node* n) {
    for (arena_group* g = a->groups; g; g = g->next_group) {




    }
    return nullptr;
}

QUESTION DATAREP-13E. Chimamanda doesn’t like that the node_find_group function from part D takes O(G) time, where G is the number of allocated arena_groups. She remembers a library function that might help, posix_memalign:

int posix_memalign(void** memptr, size_t alignment, size_t size);

The function posix_memalign() allocates size bytes and places the address of the allocated memory in *memptr. The address of the allocated memory will be a multiple of alignment, which must be a power of two and a multiple of sizeof(void*). ...

“Cool,” she says, “I can use this to speed up node_find_group!” She now allocates a new group with the following code:

arena_group* g;
int r = posix_memalign(&g, 32768, sizeof(arena_group));
assert(r == 0); // posix_memalign succeeded

Given this allocation strategy, write a version of node_find_group that takes O(1) time.

arena_group* node_find_group(arena* a, node* n) {




}

DATAREP-14. Data representation

Sort the following expressions in ascending order by value, using the operators <, =, >. For example, if we gave you:

  1. int A = 6;
  2. int B = 0x6;
  3. int C = 3;

you would write C < A = B.

  1. unsigned char a = 0x191;
  2. char b = 0x293;
  3. unsigned long c = 0xFFFFFFFF;
  4. int d = 0xFFFFFFFF;
  5. int e = d + 3;
  6. f = 4 GB
  7. size_t g = sizeof(*s) (given short *s)
  8. long h = 256;
  9. i = 0b100000000000000000000000000000000000 (binary)
  10. unsigned long j = 0xACE - 0x101;

DATAREP-15. Memory

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 starting at address 0x08048000
  6. in a read/write data segment
  7. in a register

Assume the following code, compiled without optimization.

#include <stdio.h>
#include <stdlib.h>
const long maxitems = 1000;
struct info {
    char name[20];
    unsigned int age;
    short height;
} s = { "sushi", 1, 9 };

int main(int argc, char* argv[]) {
    static long L = 0xbadf00d;
    unsigned long u = 0x8badf00d;
    int i, num = maxitems + 1;
    struct info *sp;
    printf("What did you do? %lx?\n", u);
    while (num > maxitems || num < 10) {
        printf("How much of it did you eat? ");
        scanf(" %d", &num);
    }
    sp = (struct info *)malloc(num * sizeof(*sp));
    for (i = 0; i < num; i++) {
        sp[i] = s;
    }
    return 0xdeadbeef;
}

QUESTION DATAREP-15A. The value 0xdeadbeef, when we are returning from main.

QUESTION DATAREP-15B. The variable maxitems

QUESTION DATAREP-15C. The structure s

QUESTION DATAREP-15D. The structure at sp[9]

QUESTION DATAREP-15E. The variable u

QUESTION DATAREP-15F. main

QUESTION DATAREP-15G. printf

QUESTION DATAREP-15H. argc

QUESTION DATAREP-15I. The number the user enters

QUESTION DATAREP-15J. The variable L

DATAREP-16. Memory and pointers

⚠️ This question may benefit from Unit 4, kernel programming. ⚠️

If multiple processes are sharing data via mmap, they may have the file mapped at different virtual addresses. In this case, pointers to the same object will have different values in the different processes. One way to store pointers in mmapped memory so that multiple processes can access them consistently is using relative pointers. Rather than storing a regular pointer, you store the offset from the beginning of the mmapped region and add that to the address of the mapping to obtain a real pointer. An alternative representation is called self-relative pointers. In this case, you store the difference in address between the current location (i.e., the location containing the pointer) and the location to which you want to point. Neither representation addresses pointers between the mmapped region and the rest of the address space; you may assume such pointers do not exist.

QUESTION DATAREP-16A. State one advantage that relative pointers have over self-relative pointers.

QUESTION DATAREP-16B. State one advantage that self-relative pointers have over relative pointers.

For the following questions, assume the following setup:

char* region; /* Address of the beginning of the region. */

// The following are sample structures you might find in
// a linked list that you are storing in an mmaped region.

struct ll1 {
     unsigned value;
     TYPE1 r_next; /* Relative Pointer. */
};
struct ll2 {
     unsigned value;
     TYPE2 sr_next; /* Self-Relative Pointer. */
};
ll1 node1;
ll2 node2;

QUESTION DATAREP-16C. Propose a type for TYPE1 and give 1 sentence why you chose that type.

QUESTION DATAREP-16D. Write a C expression to generate a (properly typed) pointer to the element referenced by the r_next field of ll1.

QUESTION DATAREP-16E. Propose a type for TYPE2 and give 1 sentence why you chose that type.

QUESTION DATAREP-16F. Write a C expression to generate a (properly typed) pointer to the element referenced by the sr_next field of ll2.

DATAREP-17. Data representation: Allocation sizes

union my_union {
    int f1[4];
    long f2[2];
};

int main() {
    void* p = malloc(sizeof(char*));
    my_union u;
    my_union* up = &u;
    ....
}

How much user-accessible space is allocated on the stack and/or the heap by each of the following statements? Assume x86-64.

QUESTION DATAREP-17A. union my_union { ... };

QUESTION DATAREP-17B. void* p = malloc(sizeof(char*));

QUESTION DATAREP-17C. my_union u;

QUESTION DATAREP-17D. my_union* up = &u;

DATAREP-18. Data representation: ENIAC

Professor Kohler has been developing Eddie’s NIfty Awesome Computer (ENIAC). When he built the C compiler for ENIAC, he assigned the following sizes and alignments to C’s fundamental data types. (Assume that every other fundamental type has the same size and alignment as one of these.)

Type sizeof alignof
char 1 1
char* 16 16 Same for any pointer
short 4 4
int 8 8
long 16 16
long long 32 32
float 16 16
double 32 32

QUESTION DATAREP-18A. This set of sizes is valid: it obeys all the requirements set by C’s abstract machine. Give one different size assignment that would make the set as a whole invalid.

QUESTION DATAREP-18B. What alignment must the ENIAC malloc guarantee?

For the following two questions, assume the following struct on the ENIAC:

struct s {
    char f1[7];
    char *f2;
    short f3;
    int f4;
};

QUESTION DATAREP-18C. What is sizeof(struct s)?

QUESTION DATAREP-18D. What is alignof(struct s)?

The remaining questions refer to this structure definition:

// This include file defines a struct inner, but you do not know anything
// about that structure, just that it exists.
#include "inner.hh"

struct outer {
    char f1[3];
    inner f2;
    short f3;
    int f4;
};

Indicate for each statement whether the statement is always true, possibly true, or never true on the ENIAC.

QUESTION DATAREP-18E: sizeof(outer) > sizeof(inner) (Always / Possibly / Never)

QUESTION DATAREP-18F: sizeof(outer) is a multiple of sizeof(inner) (Always / Possibly / Never)

QUESTION DATAREP-18G: alignof(outer) > alignof(struct inner) (Always / Possibly / Never)

QUESTION DATAREP-18H: sizeof(outer) - sizeof(inner) < 4 (Always / Possibly / Never)

QUESTION DATAREP-18I: sizeof(outer) - sizeof(inner) > 32 (Always / Possibly / Never)

QUESTION DATAREP-18J: alignof(inner) == 2 (Always / Possibly / Never)

DATAREP-19. Undefined behavior

Which of the following expressions, instruction sequences, and code behaviors cause undefined behavior? For each question, write Defined or Undefined. (Note that the INT_MAX and UINT_MAX constants have types int and unsigned, respectively.)

QUESTION DATAREP-19A. INT_MAX + 1 (Defined / Undefined)

QUESTION DATAREP-19B. UINT_MAX + 1 (Defined / Undefined)

QUESTION DATAREP-19C.

movq $0x7FFFFFFFFFFFFFFF, %rax
addl $1, %rax

(Defined / Undefined)

QUESTION DATAREP-19D. Failed memory allocation, i.e., malloc returns nullptr (Defined / Undefined)

QUESTION DATAREP-19E. Use-after-free (Defined / Undefined)

QUESTION DATAREP-19F. Here are two functions and a global variable:

const char string[128] = ".......";
int read_nth_char(int n) {
    return string[n];
}
int f(int i) {
    if (i & 0x40) {
        return read_nth_char(i * 2);
    } else {
        return i * 2;
    }
}

C’s undefined behavior rules would allow an aggressive optimizing compiler to simplify the code generated for f. Fill in the following function with the simplest C code you can, under the constraint that an aggressive optimizing compiler might generate the same object code for f and f_simplified.

int f_simplified(int i) {





}

DATAREP-20. Bit manipulation

It’s common in systems code to need to switch data between big-endian and little-endian representations. This is because networks represent multi-byte integers using big-endian representation, whereas x86-family processors store multi-byte integers using little-endian representation.

QUESTION DATAREP-20A. Complete this function, which translates an integer from big-endian representation to little-endian representation by swapping bytes. For instance, big_to_little(0x01020304) should return 0x04030201. Your return statement must refer to the u.c array, and must not refer to x. This function is compiled on x86-64 Linux (as every function is unless we say otherwise).

unsigned big_to_little(unsigned x) {
    union {
        unsigned intval;
        unsigned char c[4];
    } u;
    u.intval = x;



    return ______________________________________;
}

QUESTION DATAREP-20B. Complete the function again, but this time write a single expression that refers to x (you may refer to x multiple times, of course).

unsigned big_to_little(unsigned x) {



    return ______________________________________;
}

QUESTION DATAREP-20C. Now write the function little_to_big, which will translate a little-endian integer into big-endian representation. You may introduce helper variables or even call big_to_little if that’s helpful.

unsigned little_to_big(unsigned x) { 






}

DATAREP-21. 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 << i) for some i. 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 integer representation z if and only if (bit(Xi) & z) != 0.

QUESTION DATAREP-21A. What is the maximum number of set elements that can be represented in a single unsigned variable on an x86 machine?

QUESTION DATAREP-21B. Match each set operation with the C operator(s) that could implement that operation. (Complement is a unary operation.)
intersection          ==
equality ~
complement &
union ^
toggle membership
(flip whether an element is in the set)		 |

QUESTION DATAREP-21C. 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) {





}

QUESTION DATAREP-21D. 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).

Expression e Integer value Represented set
0 0 {}
a == a ______________ {A}
(unsigned) ~a < (unsigned) a ______________ {A}
a < 0 ______________ ______________
(1 << (s/2)) - 1 ______________ {A,B,C,D}
a * a ______________ {C}
abs(a) ______________ ______________
x & (x - 1) ______________ {}
x - 1 ______________ {A,D}
x ______________ ______________
s ______________ ______________

DATAREP-22. Bit Tac Toe

Brenda Bitdiddle is implementing tic-tac-toe using bitwise arithmetic. (If you’re unfamiliar with tic-tac-toe, see below.) Her implementation starts like this:

struct tictactoe {
    unsigned moves[2];
};
#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.

Tic-tac-toe, also known as noughts and crosses, is a simple paper-and-pencil game for two players, X and O. The board is a 3x3 grid. The players take turns writing their symbol (X or O) in an empty square on the grid. The game is won when one player gets their symbol in all three squares in one of the rows, one of the columns, or one of the two diagonals. X goes first; played perfectly, the game always ends in a draw.

You may access the Wikipedia page for tic-tac-toe.

QUESTION DATAREP-22A. 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.

QUESTION DATAREP-22B. Complete the following function. You may use the following helper function:

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) {






}

QUESTION DATAREP-22C. 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).

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 DATAREP-22D. Repeat part A 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.

QUESTION DATAREP-22E. Repeat part B for Lydia’s values: Use popcount to complete tictactoe_nmoves.

int tictactoe_nmoves(const tictactoe* b) {








}

QUESTION DATAREP-22F. 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);







}

DATAREP-23. 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 DATAREP-23A. Write a function to find an item in the list. The function's prototype is:

ll_node* find_element(void* mapped_file, 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 nullptr if the value is not in the list.

DATAREP-24. Integer representation

Write the value of the variable or expression in each problem, using signed decimal representation.

For example, if we gave you:

  1. int i = 0xA;
  2. int j = 0xFFFFFFFF;

you would write A) 10 B) -1.

QUESTION DATAREP-24A. int i = 0xFFFF; (You may write this either in decimal or as an expression using a power of 2)

QUESTION DATAREP-24B. short s = 0xFFFF; (You may write this either in decimal or as an expression using a power of 2)

QUESTION DATAREP-24C. unsigned u = 1 << 10;

QUESTION DATAREP-24D. ⚠️ From WeensyOS: unsigned long l = PTE_P | PTE_U;

QUESTION DATAREP-24E. int j = ~0;

QUESTION DATAREP-24F. ⚠️ From WeensyOS: sizeof(x86_64_pagetable);

QUESTION DATAREP-24G. Given this structure:

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

This expression: sizeof(ps);

QUESTION DATAREP-24H. Using the structure above: sizeof(*ps);

QUESTION DATAREP-24I. unsigned char u = 0xABC;

QUESTION DATAREP-24J. signed char c = 0xABC;

DATAREP-25. Data representation

In gdb, you observe the following values for a set of memory locations.

0x100001020:   0xa0   0xb1   0xc2   0xd3    0xe4   0xf5   0x06   0x17
0x100001028:   0x28   0x39   0x4a   0x5b    0x6c   0x7d   0x8e   0x9f
0x100001030:   0x89   0x7a   0x6b   0x5c    0x4d   0x3e   0x2f   0x10
0x100001038:   0x01   0xf2   0xe3   0xd4    0xc5   0xb6   0xa7   0x96

For each C expression below, write its value in hexadecimal. For example, if we gave you:

char *cp = (char*) 0x100001020; cp[0] =

the answer would be 0xa0.

Assume the following structure and union declarations and variable definitions.

struct _s1 {
       int i;
       long l;
       short s;
};

struct _s2 {
       char c[4];
       int i;
       struct _s1 s;
};

union _u {
       char c[8];
       int i;
       long l;
       short s;
};

char* cp = (char*) 0x100001020;
struct _s1* s1 = (struct _s1*) 0x100001020;
struct _s2* s2 = (struct _s2*) 0x100001020;
union _u* u = (union _u*) 0x100001020;

QUESTION DATAREP-25A. cp[4] =

QUESTION DATAREP-25B. cp + 7 =

QUESTION DATAREP-25C. s1 + 1 =

QUESTION DATAREP-25D. s1->i =

QUESTION DATAREP-25E. sizeof(s1) =

QUESTION DATAREP-25F. &s2->s =

QUESTION DATAREP-25G. &u->s =

QUESTION DATAREP-25H. s1->l =

QUESTION DATAREP-25I. s2->s.s =

QUESTION DATAREP-25J. u->l =

DATAREP-26. Sizes and alignments

Here’s a test struct with n members. Assume an x86-64 machine, where each Ti either is a basic x86-64 type (e.g., int, char, double) or is a type derived from such types (e.g., arrays, structs, pointers, unions, possibly recursively), and assume that ai≤8 for all i.

struct test {
    T1 m1;      // sizeof(T1) == s1, alignof(T1) == a1
    T2 m2;      // sizeof(T2) == s2, alignof(T2) == a2
    ...
    Tn mn;      // sizeof(Tn) == sn, alignof(Tn) == an
};

In these questions, you will compare this struct with other structs that have the same members, but in other orders.

QUESTION DATAREP-26A. True or false: The size of struct test is minimized when its members are sorted by size. In other words, if s1≤s2≤…≤sn, then sizeof(struct test) is less than or equal to the struct size for any other member order.

If true, briefly explain your answer; if false, give a counterexample (i.e., concrete types for T1, …, Tn that do not minimize sizeof(struct test)).

QUESTION DATAREP-26B. True or false: The size of struct test is minimized when its members are sorted by alignment. In other words, if a1≤a2≤…≤an, then sizeof(struct test) is less than or equal to the struct size for any other member order.

If true, briefly explain your answer; if false, give a counterexample.

QUESTION DATAREP-26C. True or false: The alignment of struct test is minimized when its members are sorted in increasing order by alignment. In other words, if a1≤a2≤…≤an, then alignof(struct test) is less than or equal to the struct alignment for any other member order.

If true, briefly explain your answer; if false, give a counterexample.

QUESTION DATAREP-26D. What is the maximum number of bytes of padding that struct test could contain for a given n? The answer will be a pretty simple formula involving n. (Remember that ai≤8 for all i.)

QUESTION DATAREP-26E. What is the minimum number of bytes of padding that struct test could contain for a given n?

DATAREP-27. Undefined behavior

QUESTION DATAREP-27A. Sometimes a conforming C compiler can assume that a + 1 > a, and sometimes it can’t. For each type below, consider this expression:

a + (int) 1 > a

and say whether the compiler:

  1. int a
  2. unsigned a
  3. char* a
  4. unsigned char a
  5. struct {int m;} a

QUESTION DATAREP-27B. The following code checks its arguments for sanity, but not well: each check can cause undefined behavior.

void sanity_check(int* array, size_t array_size, int* ptr_into_array) {
    if (array + array_size < array) {
        fprintf(stderr, "`array` is so big that it wraps around!\n");
        abort();
    }
    if (ptr_into_array < array || ptr_into_array > array + array_size) {
        fprintf(stderr, "`ptr_into_array` doesn’t point into the array!\n");
        abort();
    }
    ...

Rewrite these checks to avoid all undefined behavior. You will likely add one or more casts to uintptr_t. For full credit, write each check as a single comparison (no && or ||, even though the current ptr_into_array check uses ||).

array_size check:

ptr_into_array check:

QUESTION DATAREP-27C. In lecture, we discussed several ways to tell if a signed integer x is negative. One of them was the following:

int isnegative = (x & (1UL << (sizeof(x) * CHAR_BIT))) != 0;

But this is incorrect: it has undefined behavior. Correct it by adding two characters.

DATAREP-28. Memory errors and garbage collection

⚠️ We didn’t discuss garbage collectors in class this year. ⚠️

Recall that a conservative garbage collector is a program that can automatically free dynamically-allocated memory by detecting when that memory is no longer referenced. Such a GC works by scanning memory for currently-referenced pointers, starting from stack and global memory, and recursing over each referenced object until all referenced memory has been scanned. We built a conservative garbage collector in lecture datarep6.

QUESTION DATAREP-28A. An application program that uses conservative GC, and does not call free directly, will avoid certain errors and undefined behaviors. Which of the following errors are avoided? List all that apply.

  1. Use-after-free
  2. Double free
  3. Signed integer overflow
  4. Boundary write error
  5. Unaligned access

QUESTION DATAREP-28B. Write a C program that leaks unbounded memory without GC, but does not do so with GC. You should need less than 5 lines. (Leaking “unbounded” memory means the program will exhaust the memory capacity of any machine on which it runs.)

QUESTION DATAREP-28C. Not every valid C program works with a conservative GC, because the C abstract machine allows a program to manipulate pointers in strange ways. Which of the following pointer manipulations might cause the conservative GC from class to inappropriately free a memory allocation? List all that apply.

  1. Storing the pointer in a uintptr_t variable

  2. Writing the pointer to a disk file and reading it back later

  3. Using the least-significant bit of the pointer to store a flag:

    int* set_ptrflag(int* x, int flagval) {
    return (int*) ((uintptr_t) x | (flagval ? 1 : 0));
}
int get_ptrflag(int* x) {
    return (uintptr_t) x & 1;
}
int deref_ptrflag(int* x) {
    return *((int*) ((uintptr_t) x & ~1UL));
}

  4. Storing the pointer in textual form:

    void save_ptr(char buf[100], void* p) {
    sprintf(buf, "%p", p);
}
void* restore_ptr(const char buf[100]) {
    void* p;
    sscanf(buf, "%p", &p);
    return p;
}

  5. Splitting the pointer into two parts and storing the parts in an array:

    typedef union {
    unsigned long ival;
    unsigned arr[2];
} value;
value save_ptr(void* p) {
    value v;
    v.arr[0] = (uintptr_t) p & 0xFFFFFFFFUL;
    v.arr[1] = ((uintptr_t) p / 4294967296UL) & 0xFFFFFFFFUL;
    return v;
}

DATAREP-29. Bitwise

QUESTION DATAREP-29A. Consider this C fragment:

uintptr_t x = ...;
uintptr_t r = 0;
if (a < b) {
    r = x;
}

Or, shorter:

uintptr_t r = a < b ? x : 0;

Write a single expression that evaluates to the same value, but that does not use the conditional ?: operator. You will use the fact that a < b always equals 0 or 1. For full credit, do not use expensive operators (multiply, divide, modulus).

QUESTION DATAREP-29B. This function returns one more than the index of the least-significant 1 bit in its argument, or 0 if its argument is zero.

int ffs(unsigned long x) {
    for (int i = 0; i < sizeof(x) * CHAR_BIT; ++i) {
        if (x & (1UL << i)) {
            return i + 1;
        }
    }
    return 0;
}

This function runs in O(B) time, where B is the number of bits in an unsigned long. Write a version of ffs that runs instead in O(log B) time.

DATAREP-30. Data representation

QUESTION DATAREP-30A. Write a type whose size is 19,404,329 times larger than its alignment.

QUESTION DATAREP-30B. Consider a structure type T with N members, all of which have nonzero size. Assume that sizeof(T) == alignof(T). What is N?

QUESTION DATAREP-30C. What is a C type that obeys (T) -1 == (T) 255 on x86-64?

Parts D–G use this code. The architecture might or might not be x86-64.

unsigned char a[] = {
    0x7A, 0xEC, 0x0D, 0xBE, 0x99, 0x0A, 0xD8, 0x0E
};
unsigned* s1 = (unsigned*) &a[0];
unsigned* s2 = s1 + 1;

Assume that (uintptr_t) s2 - (uintptr_t) s1 == 4 and *s1 > *s2.

QUESTION DATAREP-30D. What is sizeof(a)?

QUESTION DATAREP-30E. What is sizeof(unsigned) on this architecture?

QUESTION DATAREP-30F. Is this architecture big-endian or little-endian?

QUESTION DATAREP-30G. Might the architecture be x86-64?

DATAREP-31. Memory errors

Mavis Gallant is starting on her debugging memory allocator. She’s written code that aims to detect invalid frees, where a pointer passed to m61_free was not returned by an earlier m61_malloc.

D1.    typedef struct m61_metadata {
D2.        size_t magic;
D3.        size_t padding;
D4.    } m61_metadata;

M1.    void* m61_malloc(size_t sz) {
M2.        m61_metadata* meta = base_malloc(sz + sizeof(m61_metadata));
M3.        meta->magic = 0x84157893401;
M4.        return (void*) (meta + 1);
M5.    }

F1.    void m61_free(void* ptr) {
F2.        m61_metadata* meta = (m61_metadata*) ptr - 1;
F3.        if (meta->magic != 0x84157893401) {
F4.            fprintf(stderr, "invalid free of %p\n", ptr);
F5.            abort();
F6.        }
F7.        base_free(ptr);
F8.    }

C1.    void* m61_calloc(size_t count, size_t sz) {
C2.        void* p = m61_malloc(sz * count);
C3.        memset(p, 0, sz * count);
C4.        return p;
C5.    }`

Help her track down bugs.

QUESTION DATAREP-31A. What is sizeof(struct m61_metadata)?

QUESTION DATAREP-31B. Give an m61_ function call (function name and arguments) that would cause both unsigned integer overflow and invalid memory accesses.

QUESTION DATAREP-31C. Give an m61_ function call (function name and arguments) that would cause integer overflow, but no invalid memory access within the m61_ functions. (The application might or might not make an invalid memory access later.)

QUESTION DATAREP-31D. These functions have some potential null pointer dereferences. Fix one such problem, including the line number(s) where your code should go.

QUESTION DATAREP-31E. Put a single line of C code in the blank. The resulting program should (1) be well-defined with no memory leaks when using default malloc/free/calloc, but (2) always cause undefined behavior when using Mavis’s debugging malloc/free/calloc.

... #includes ...
int main(void) {



    ________________________
}

QUESTION DATAREP-31F. A double free should print a different message than an invalid free. Write code so Mavis’s implementation does this; include the line numbers where the code should go.