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Computer Science 61 and E61
Systems Programming and Machine Organization
This is the 2013 version of the course. Main site

Sample Final Questions

The final will be cumulative, though it will be weighted more towards the second half of the class. So why not check out:

1. Expressions

QUESTION 1A. 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 machine.

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

2. Hello program

Recall from the sample midterm questions that Shintaro Tsuji is representing images such as this one in computer memory:


(Zoomed in:)


He stores each image in an array of 16-bit unsigned integers:

uint16_t cute[16];

Row Y of the image is stored in integer cute[Y]. Given this structure, Shintaro 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:


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

Hello-f0.gif   Hello-f7.gif   Hello-f2.gif   Hello-f3.gif   Hello-f4.gif
Hello-f5.gif   Hello-f6.gif   Hello-f1.gif   Hello-f8.gif   Hello-f9.gif

For each function, what image does that function create?


void f1() {
   for (int i = 0; i < 16; ++i)
      cute[i] = ~cute[i];


void f2() {
   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);


void f3() {
   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 f4() {
   for (int i = 0; i < 16; ++i)
      cute[i] &= ~(7 << i);
void f5() {
   for (int i = 0; i < 16; ++i)
      cute[i] <<= i/4;
void f6() {
   for (int i = 0; i < 16; ++i)
      cute[i] = -1 * !!(cute[i] & 64);
void f7() {
   for (int i = 0; i < 16; ++i) {
      int tmp = cute[15-i];
      cute[15-i] = cute[i];
      cute[i] = tmp;
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] ^ cute[i];
void f10() {
   for (int i = 0; i < 16; ++i)
      cute[i] = cute[i] ^ 4080;

3. Data structures

Here are four assembly functions, f1 through f4.

	movl	4(%esp), %eax
	movl	8(%esp), %ecx
	testl	%ecx, %ecx
	jle	.L2
	xorl	%edx, %edx
	movl	4(%eax), %eax
	incl	%edx
	cmpl	%ecx, %edx
	jne	.L3
	movl	(%edx), %eax
	movl	8(%esp), %edx
	leal	0(,%edx,4), %ecx
	movl	4(%esp), %eax
	movl	(%eax,%ecx), %eax
	addl	%ecx, %eax
	movl	(%eax), %eax
	pushl	%esi
	pushl	%ebx
	movl	12(%esp), %ecx
	movl	16(%esp), %esi
	movl	20(%esp), %eax
	testl	%esi, %esi
	jle	.L9
	xorl	%edx, %edx
	movl	%eax, %ebx
	andl	$1, %ebx
	movl	4(%ecx,%ebx,4), %ecx
	incl	%edx
	sarl	%eax
	cmpl	%esi, %edx
	jne	.L10
	movl	(%ecx), %eax
	popl	%ebx
	popl	%esi
	movl	8(%esp), %edx
	movl	4(%esp), %eax
	movl	(%eax,%edx,4), %eax

QUESTION 3A. 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 3B. 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

4. Disassemble (14 points)

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

	.string	"%d %d\n"

	.globl	f
	.type	f, @function
	pushl	%ebp
	movl	$1, %ecx
	movl	%esp, %ebp
	pushl	%edi
	pushl	%esi
	pushl	%ebx
	subl	$12, %esp
	movl	$1, %eax
	movl	%eax, %ebx
	imull	%eax, %ebx
	movl	%ecx, %esi
	imull	%ecx, %esi
	movl	$1, %edx
	movl	%edx, %edi
	imull	%edx, %edi
	addl	%ebx, %edi
	cmpl	%esi, %edi
	je	.L11
	incl	%edx
	cmpl	%eax, %edx
	jle	.L4
	incl	%eax
	cmpl	%ecx, %eax
	jle	.L9
	incl	%ecx
	jmp	.L13
	pushl	%ecx
	pushl	%eax
	pushl	%edx
	pushl	$.LC1
	call	printf
	leal	-12(%ebp), %esp
	movl	$1, %eax
	popl	%ebx
	popl	%esi
	popl	%edi
	popl	%ebp

QUESTION 4A. How many arguments might this function have? Circle all that apply.

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

QUESTION 4B. 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 4C. Which callee-saved registers does this function save and restore? Circle all that apply.

  1.  %eax
  2.  %ebx
  3.  %ecx
  4.  %edx
  5.  %ebp
  6.  %esi
  7.  %edi
  8. None of the above

QUESTION 4D. 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

QUESTION 4E. 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

5. Machine programming (20 points)

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.

1. cmpl %ecx, %edx
2. decl %edx
3. incl %eax
4. je L1
5. jl L2
6. jmp L3
7. movl 4(%esp), %ecx [movc]
8. movl 8(%esp), %edx [movd]
9. movl (%ecx,%eax,4), %ecx [movx]
10. ret
11. xchgl %eax, %ecx
12. xorl %eax, %eax

(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:


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.


 int return_zero(void) {
     return 0;


 int identity(int a) {
     return a;


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


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

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 5E. [Reminder: Complete at least one of 5E–5G for full credit.]

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

QUESTION 5F. [Reminder: Complete at least one of 5E–5G for full credit.]

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

QUESTION 5G. [Reminder: Complete at least one of 5E–5G for full credit.]

 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];
             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.)

6. Virtual memory (15 points)

QUESTION 6A. 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 WeensyOS (Assignment 3) 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 6B. What is the maximum size (in pages) of an x86 page table?

QUESTION 6C. What is the minimum size (in pages) of an x86 page table that would allow a process to access 222 distinct physical addresses?

The x86 architecture we discussed in class 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 6D. 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 6E. 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.

7. Cost expressions (15 points)

In the following questions, you will reason about the abstract costs of various operations, using the following tables of constants.

Table of Basic Costs

S System call overhead (i.e., entering and exiting the kernel)
F Page fault cost (i.e., entering and exiting the kernel)
P Cost of allocating a new physical page
M Cost of installing a new page mapping
B Cost of copying a byte

Table of Sizes

nk Number of memory pages allocated to the kernel
Per-process sizes (defined for each process p)
np Number of memory pages allocated to p
rp Number of read-only memory pages allocated to p
wp = nprp Number of writable memory pages allocated to p
mp Number of memory pages actually modified by p after the previous fork()

One of our tiny operating systems from class (OS02) included a program that called a recursive function. When the recursive function’s argument grew large enough, the stack pointer moved beyond the memory actually allocated for the stack, and the program crashed.

QUESTION 7A. In our first solution for this problem, the process called the sys_page_alloc(void *addr) system call, which allocated and mapped a single new page at address addr (the new stack page). Write an expression for the cost of this sys_page_alloc() system call in terms of the constants above.

QUESTION 7B. Our second solution for this problem changed the operating system’s page fault handler. When a fault occurred in a process’s stack region, the operating system allocated a new page to cover the corresponding address and restarted the process. Write an expression for the cost of such a fault in terms of the constants above.

QUESTION 7C. Design a revised version of sys_page_alloc that supports batching. Give its signature and describe its behavior.

QUESTION 7D. Write an expression for the cost of a call to your batching allocation API.

In the remaining questions, a process p calls fork(), which creates a child process, c.

Assume that the base cost of performing a fork() system call is Φ. This cost includes the fork() system call overhead (S), the overhead of allocating a new process, the overhead of allocating a new page directory with kernel mappings, and the overhead of copying registers. But it does not include overhead from allocating, copying, or mapping other memory.

QUESTION 7E. Consider the following implementations of fork():

A. Naive fork: Copy all process memory (Assignment 3, Step 5).
B. Eager fork: Copy all writable process memory; share read-only process memory, such as code (Assignment 3, Step 6).
C. Copy-on-write fork: initially share all memory as read-only. Create writable copies later, on demand, in response to write faults (Assignment 3 extra credit).

Which expression best represents the total cost of the fork() system call in process p, for each of these fork implementations? Only consider the system call itself, not later copy-on-write faults.

(Note: Per-process variables, such as n, are defined for each process. So, for example, np is the number of pages allocated to the parent process p, and nc is the number of pages allocated to the child process c.)

  1. Φ
  2. Φ + np × M
  3. Φ + (np + wp) × M
  4. Φ + np × 212 × (B + F)
  5. Φ + np × (212B + P + M)
  6. Φ + np × (P + M)
  7. Φ + wp × (212B + P + M)
  8. Φ + np × (212B + P + M) − rp × (212B + P)
  9. Φ + np × M + mc × (P + F)
  10. Φ + np × M + mc × (212B + F + P)
  11. Φ + np × M + (mp+mc) × (P + F)
  12. Φ + np × M + (mp+mc) × (212B + F + P)

QUESTION 7F. When would copy-on-write fork be more efficient than eager fork (meaning that the sum of all fork-related overheads, including faults for pages that were copied on write, would be less for copy-on-write fork than eager fork)? Circle the best answer.

  1. When np < nk.
  2. When wp × F < wp × (M + P).
  3. When mc × (F + M + P) < wp × (M + P).
  4. When (mp+mc) × (F + M + P + 212B) < wp × (P + 212B).
  5. When (mp+mc) × (F + P + 212B) < wp × (P + M + 212B).
  6. When mp < mc.
  7. None of the above.

8. Networking (10 points)

QUESTION 8A. Which of the following system calls should a programmer expect to sometimes block (i.e., to return after significant delay)? Circle all that apply.

1. socket 5. connect
2. read 6. write
3. accept 7. usleep
4. listen 8. None of these

QUESTION 8B. 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.

1. Loss 4. Duplication 7. Exponential backoff
2. Delay 5. Batching
3. Reordering 6. Prefetching
Finalfig2012 1.gif Finalfig2012 2.gif Finalfig2012 3.gif Finalfig2012 4.gif
Finalfig2012 5.gif Finalfig2012 6.gif Finalfig2012 7.gif

9. Threads (15 points)

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 9A. What is wrong? Describe why the problem is a synchronization issue.

QUESTION 9B. 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.

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 9C. What is the new problem? Describe why it is a synchronization issue.

QUESTION 9D. Write pseudocode or C code that fixes this problem. You should refer to pthread functions. For full credit your solution should have low contention.

10. Processes

So that's how to use existing system calls to build a new abstraction. But we can also build 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.

int result; int result;
result = rendezvous(0, 5); printf("About to rendezvous\n");
result = rendezvous(0, 600);
/* The processes swap data; both become runnable */
printf("Process A got %d\n", result); printf("Process B got %d\n", result);

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 10A. 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 10B. Can a rendezvous-pipe support all pipe features?