CS代考 IA32 code: – cscodehelp代写
Instructions:
D (print clearly!): Full Name:
15-213/18-213, Fall 2011 Final Exam
Friday, December 16, 2011
Copyright By cscodehelp代写 加微信 cscodehelp
• Make sure that your exam is not missing any sheets, then write your D and full name on the front.
• This exam is closed book, closed notes (except for 2 double-sided note sheets). You may not use any electronic devices.
• Write your answers in the space provided below the problem. If you make a mess, clearly indicate your final answer.
• The exam has a maximum score of 92 points.
• The problems are of varying difficulty. The point value of each problem is indicated. Good luck!
TOTAL (92):
Page 1 of 17
Problem 1. (10 points):
General systems topics. Write your answer for each question in the following table: 1 2 3 4 5 6 7 8 9 10
1. Consider a direct-mapped cache memory. Which one of the following statements is true?
(a) The cache has 1 line per set.
(b) The cache has 1 word per block. (c) The cache has 1 set per cache. (d) None of the above.
2. Which one of the following statements about cache memories is true:
(a) Larger caches are more susceptible to capacity misses than smaller caches.
(b) Caches with lower associativity are more susceptible to conflict misses than those with higher associativy.
(c) Caches with higher associativity are more susceptile to cold misses than those with lower asso- ciativey.
(d) None of the above
3. Which one of the following is NOT contained in an ELF executable file?
(a) Machine code (b) Global variables (c) User stack
(d) Symbol table
4. Assuming no errors, which one of the following statements about fork is true?
(a) Called once, returns once. (b) Called once, returns twice. (c) Called once, returns never. (d) Called twice, returns once. (e) None of the above.
5. Assuming no errors, which one of the following statements about execve is true?
(a) Called once, returns once. (b) Called once, returns twice. (c) Called once, returns never. (d) Called twice, returns once. (e) None of the above.
Page 2 of 17
6. Which one of the following statements about processes is false?
(a) The operating system kernel runs as its own separate process. (b) Each process shares the CPU with other processes.
(c) Each process has its own private address space.
(d) The environment for a process is stored on the stack.
7. What happens if the parent of a zombie child terminates?
(a) The zombie child becomes a wraith and is never reaped.
(b) The zombie child is reaped by the init process.
(c) The zombie child is reaped by the process with the nearest PID. (d) None of the above.
8. SupposethatthekerneldeliverstwoSIGCHLDsignalstotheparentwhiletheparentisnotscheduled. When the kernel finally schedules the parent, how many times will the SIGCHLD handler be called?
(a) None, because sending multiple signals will always crash the program. (b) Exactly once, because signals are not queued.
(c) Exactly twice, because signals are queued.
(d) More than twice, depending on how the handler is installed.
9. Which one of the following statements is NOT true of storage allocators?
(a) In the best case, coalescing with boundary tags is linear in the number of free blocks. (b) Seglists typically approximate best fit search.
(c) Payloads must be aligned to some boundary.
(d) Explicit lists are typically faster than implicit lists.
(e) None of the above.
10. Which one of the following addresses is 8-byte aligned?
(a) 11101101011101112 (b) 11101101011101002 (c) 11101101011100002 (d) 11101101011101102 (e) None of the above
Page 3 of 17
Problem 2. (8 points):
Floating point encoding. Consider the following 5-bit floating point representation based on the IEEE floating point format. This format does not have a sign bit – it can only represent nonnegative numbers.
• There are k = 3 exponent bits. The exponent bias is 3.
• There are n = 2 fraction bits.
Recall that numeric values are encoded as a value of the form V = M × 2E , where E is the exponent after biasing, and M is the significand value. The fraction bits encode the significand value M using either a denormalized (exponent field 0) or a normalized representation (exponent field nonzero). The exponent E is given by E = 1 − Bias for denormalized values and E = e − Bias for normalized values, where e is the value of the exponent field exp interpreted as an unsigned number.
Below, you are given some decimal values, and your task it to encode them in floating point format. In addition, you should give the rounded value of the encoded floating point number. To get credit, you must give these as whole numbers (e.g., 17) or as fractions in reduced form (e.g., 3/4). Any rounding of the significand is based on round-to-even, which rounds an unrepresentable value that lies halfway between two representable values to the nearest even representable value.
9/32 2 13 1/8 15/4
Floating Point Bits
Rounded value 1/4
Page 4 of 17
Problem 3. (6 points):
Array indexing. Consider the C code below, where H and J are constants declared with #define. int array1[H][J];
int array2[J][H];
void copy_array(int x, int y) {
array2[y][x] = array1[x][y];
Suppose the above C code generates the following x86-64 assembly code:
# On entry:
# %edi = x
# %esi = y
copy_array:
movslq %edi,%rdi
movslq %esi,%rsi
movq %rsi, %rdx
salq $4, %rdx
subq %rsi, %rdx
addq %rdi, %rdx
leaq 0(,%rdi,8), %rax
subq %rdi, %rax
addq %rsi, %rax
movl array1(,%rax,4), %eax
movl %eax, array2(,%rdx,4)
What are the values of H and J?
Page 5 of 17
Problem 4. (10 points):
Structure access. Consider the following data structure declarations:
struct node {
struct data d;
struct node *next;
Below are given four C functions and four x86-64 code blocks. Next to each of the x86-64 code blocks, write the name of the C function that it implements.
struct data {
char str[16];
int alpha(struct node *ptr) {
return ptr->d.x;
char *beta(struct node *ptr) {
ptr = ptr->next;
return ptr->d.str;
char gamma(struct node *ptr) {
return ptr->d.str[7];
long *delta(struct node *ptr) {
struct data *dp =
(struct data *) ptr;
return &dp->x;
char *epsilon(struct node *ptr) {
return &ptr->d.str[2];
movsbl 15(%rdi),%eax
movq (%rdi), %rax
movq 24(%rdi), %rax
addq $8, %rax
movq %rdi, %rax
leaq 10(%rdi), %rax
Page 6 of 17
Problem 5. (6 points):
Loops. Consider the following x86-64 assembly function: loopy:
# a in %rdi, n in %esi
movl $0, %ecx
movl $0, %edx
testl %esi, %esi
movslq %edx,%rax
movl (%rdi,%rax,4), %eax
cmpl %eax, %ecx
cmovl %eax, %ecx
addl $1, %edx
cmpl %ecx, %esi
jg .L6
movl %ecx, %eax
Fill in the blanks of the corresponding C code.
• You may only use the C variable names n, a, i and x, not register names. • Use array notation in showing accesses or updates to elements of a.
int loopy(int a[], int n)
int x = _____;
for(i = ____________; ____________; ____________) {
if (____________)
return x; }
x = ____________;
Page 7 of 17
Problem 6. (12 points):
Stack discipline.
A. (2 pts) Consider the following snippet of IA32 code:
8048390: call 8048395
8048395: pop %eax
Suppose that just before the call instruction executes, %esp = 0xffffd834. Then what is the value of %eax after the pop instruction executes?
%eax = 0x_________________
B. (2 pts) Consider a slightly different snippet of IA32 code:
8048396: call 804839b
804839b: ret
Suppose that just before the call instruction executes, %esp = 0xffffd838. Then what is the value of %eip after the ret instruction executes?
%eip = 0x_________________
(Please go to the next page for Parts C and D)
Page 8 of 17
Stack discipline (continued)
C. (7 points) Consider the following bit of C code and its dissassembled IA32 machine code (notice the header comment):
int power(int x, int n)
804839c: leave
804839d: ret
else return
int main() {
power(x, n-1) * x;
push %ebp
mov %esp,%ebp
sub $0x8,%esp
mov 0xc(%ebp),%edx
mov $0x1,%eax
test %edx,%edx
je 804839c
mov %eax,0x4(%esp)
mov 0x8(%ebp),%eax
mov %eax,(%esp)
call 8048374
08048374
# On entry to power(2,4):
# %esp = 0xffffd81c, %ebp = 0xffffd838
Suppose that the main routine calls power(2,4) from some unspecified location. Fill in the values on the stack immediately after the subsequent call to power(2,3). If a value is unknown, please write UNKNOWN in the blank:
Address Value
0xffffd824 0x___________________ 0xffffd820 0x___________________ 0xffffd81c 0x___________________ 0xffffd818 0x___________________ 0xffffd814 0x___________________ 0xffffd810 0x___________________ 0xffffd80c 0x___________________
D. (1 point) What is the value of %ebp immediately after the call to power(2,3): %ebp = 0x___________________
Page 9 of 17
Problem 7. (6 points):
Caches. In this problem you will estimate the miss rates for some C functions. Assumptions:
• 16-way set associative L1 cache (E = 16) with a block size of 32 bytes (B = 32). • N is very large, so that a single row or column cannot fit in the cache.
• sizeof(int) == 4
• Variables i, k, and sum are stored in registers.
• The cache is cold before each function is called. Part A (3 points)
int sum1(int A[N][N], int B[N][N])
int i, k, sum = 0;
for (i = 0; i < N; i++)
for (k = 0; k < N; k++)
sum += A[i][k] + B[k][i];
return sum;
Part B (3 points)
int sum2(int A[N][N], int B[N][N])
int i, k, sum = 0;
for (i = 0; i < N; i++)
for (k = 0; k < N; k++)
sum += A[i][k] + B[i][k];
return sum;
Circle the closest miss rate for sum1: • 1/16
• 1/8 • 1/4 • 1/2 • 9/16 •1
Circle the closest miss rate for sum2: • 1/16
• 1/8 • 1/4 • 1/2 • 9/16 •1
Page 10 of 17
Problem 8. (10 points):
Exceptional control flow.
A. Consider the following C program. Assume the program executes to completion and that fork, waitpid, and printf always succeed.
int main() {
pid_t pid;
int sum = 0;
if ((pid = fork()) == 0)
sum += 10;
waitpid(pid, NULL, 0);
sum += 20;
if (pid > 0)
printf(“Parent: sum=%d
”, sum);
printf(“Child: sum=%d
”, sum);
return 0; }
Show the output of this program:
Child: sum=________
Parent: sum=________
B. Now consider the same program as in Part A, but with the call to waitpid removed. Assume the program executes to completion and that printf always succeeds. Make no assumptions about the results of the other function calls.
List all of the possible outputs of such a program. Each blank box holds the complete output from one execution of the program. Some blank boxes may be left unused.
+———————+———————+———————+ |||| |||| +———————+———————+———————+ |||| |||| +———————+———————+———————+
(Please goto the next page for Part C.)
Page 11 of 17
C. Consider the C program below. Assume the program runs to completion and that all functions return normally.
int main () {
if (fork() == 0) {
if (fork() == 0) {
printf(“9”);
exit(1); }
printf(“5”);
pid_t pid;
if ((pid = wait(NULL)) > 0) {
printf(“3”);
printf(“0”);
return 0; }
List four possible outputs for this program:
1. _________________
2. _________________
3. _________________
4. _________________
Page 12 of 17
Problem 9. (9 points):
Imagine a system with the following attributes:
• The system has 1MB of virtual memory
• The system has 256KB of physical memory
• The page size is 4KB
• The TLB is 2-way set associative with 8 total entries.
The contents of the TLB and the first 32 entries of the page table are given below. All numbers are in hexadecimal.
Page Table
VPN PPN Valid VPN PPN Valid
00 17 1 10 26 0 01 28 1 11 17 0 02 14 1 12 0E 1 03 0B 0 13 10 1 04 26 0 14 13 1 05 13 0 15 1B 1 06 0F 1 16 31 1 07 10 1 17 12 0 081C018231 0925119040 0A 31 0 1A 0C 1 0B 16 1 1B 2B 0 0C 01 0 1C 1E 0 0D 15 0 1D 3E 1 0E 0C 0 1E 27 1 0F 2B 1 1F 15 1
0 05 13 1 3F 15 1
1 10 0F 1 0F 1E 0
2 1F011 111F0
3 032B1 1D 23 0
Page 13 of 17
A. Warmup Questions
(a) How many bits are needed to represent the virtual address space? _____ (b) How many bits are needed to represent the physical address space? _____ (c) How many bits are needed to represent a page table offset? _____
B. Virtual Address Translation I
Please step through the following address translation. Indicate a page fault by entering ’-’ for Physical Address.
Virtual address: 0x1F213 Parameter Value
VPN 0x TLB Index 0x TLB Tag 0x
Parameter Value
TLB Hit? (Y/N)
Page Fault? (Y/N) Physical Address 0x
Use the layout below as scratch space for the virtual address bits. To allow us to give you partial credit, clearly mark the bits that correspond to the VPN, TLB index (TLBI), and TLB tag (TLBT).
19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
(Please go to the next page for part C)
Page 14 of 17
C. Virtual Address Translation II
Please step through the following address translation. Indicate a page fault by entering ’-’ for Physical Address.
Virtual address: 0x14213 Parameter Value
VPN 0x TLB Index 0x TLB Tag 0x
Parameter Value
TLB Hit? (Y/N)
Page Fault? (Y/N) Physical Address 0x
Use the layout below as scratch space for the virtual address bits. To allow us to give you partial credit, clearly mark the bits that correspond to the VPN, TLB index (TLBI), and TLB tag (TLBT).
19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0
Page 15 of 17
Problem 10. (6 points):
A. Suppose that the disk file foobar.txt consists of the six ASCII characters “foobar”. What is the output of the following program?
/* any necessary includes */
char buf[20] = {0}; /* init to all zeroes */
int main(int argc, char* argv[]) {
int fd1 = open(“foobar.txt”, O_RDONLY);
int fd2 = open(“foobar.txt”, O_RDONLY);
dup2(fd2, fd1);
read(fd1, buf, 3);
close(fd1);
read(fd2, &buf[3], 3);
close(fd2);
printf(“buf = %s
”, buf);
return 0; }
Output: buf = _______________
B. Now consider the identical program, except that dup2 is commented out. What is the output of this
char buf[20] = {0}; /* init to all zeroes */
int main(int argc, char* argv[]) {
int fd1 = open(“foobar.txt”, O_RDONLY);
int fd2 = open(“foobar.txt”, O_RDONLY);
//dup2(fd2, fd1);
read(fd1, buf, 3);
close(fd1);
read(fd2, &buf[3], 3);
close(fd2);
printf(“buf = %s
”, buf);
return 0; }
Output: buf = _______________
Page 16 of 17
Problem 11. (9 points):
Synchronization. This problem is about using semaphores to synchronize access to a shared bounded FIFO queue in a producer/consumer system with an arbitrary number of producers and consumers.
• The queue is initially empty and has a capacity of 10 data items.
• Producer threads call the insert function to insert an item onto the rear of the queue.
• Consumer threads call the remove function to remove an item from the front of the queue. • The system uses three semaphores: mutex, items, and slots.
Your task is to use P and V semaphore operations to correctly synchronize access to the queue.
A. What is the initial value of each semaphore?
mutex = _______
items = _______
slots = _______
B. Add the appropriate P and V operations to the psuedo-code for the insert and remove functions:
void insert(int item) int remove() {{
/* Insert sem ops here */
add_item(item);
/* Insert sem ops here */
/* Insert sem ops here */
item = remove_item();
/* Insert sem ops here */
return item;
Page 17 of 17
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