# CS代考 IA32 stack discipline, fill in the stack diagram with the values that would – cscodehelp代写

Instructions:
D (print clearly!): Full Name:
15-213/18-213, Fall 2011 Exam 1
Tuesday, October 18, 2011

• 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 1 double-sided note sheet). You may not use any electronic devices.
• The exam has a maximum score of 66 points.
• The problems are of varying difficulty. The point value of each problem is indicated. Good luck!
TOTAL (66):
Page 1 of 15

Problem 1. (8 points):
Multiple choice. Write your answer for each question in the following table: 12345678
1. What is the minimum (most negative) value of a 32-bit two’s complement integer?
(b) −232 + 1 (c) −231
(d) −231 + 1
2. What is the difference between the mov and lea instructions?
(a) lea dereferences an address, while mov doesn’t.
(b) mov dereferences an address, while lea doesn’t.
(c) lea can be used to copy a register into another register, while mov cannot. (d) mov can be used to copy a register into another register, while lea cannnot.
3. After executing the following code, which of the variables are equal to 0?
unsigned int a = 0xffffffff;
unsigned int b = 1;
unsigned int c = a + b;
unsigned long d = a + b;
unsigned long e = (unsigned long)a + b;
(Assume ints are 32 bits wide and longs are 64 bits wide.)
(a) None of them (b) c
(c) c and d
(d) c, d, and e
Page 2 of 15

4. Assume a function foo takes two arguments. When calling foo(arg1, arg2), which is the correct order of operations assuming x86 calling conventions and that foo must allocate stack space (implies that we must save the %ebp)?
(a) push arg1, push arg2, call foo, push %ebp (b) push arg1, push arg2, push %ebp, call foo (c) push arg2, push arg1, call foo, push %ebp (d) push arg2, push arg1, push %ebp, call foo
5. Which one of the following statements is NOT true?
(a) x86-64 provides a larger virtual address space than x86. (b) The stack disciplines for x86 and x86-64 are different.
(c) x86 uses %ebp as the base pointer for the stack frame. (d) x86-64 uses %rbp as the base pointer for the stack frame.
6. Consider a 4-way set associative cache (E = 4). Which one of the following statements it true?
(a) The cache has 4 blocks per line. (b) The cache has 4 sets per line. (c) The cache has 4 lines per set. (d) The cache has 4 sets per block. (e) None of the above.
7. Which one of the following statements about cache memories is true?
(a) Fully associative caches offer better latency, while direct-mapped caches have lower miss rates. (b) Fully associative caches offer lower miss rates, while direct-mapped caches have better latency. (c) Direct-mapped caches have both better miss rates and better latency.
(d) Both generally have similar latency and miss rates.
8. Consider an SRAM-based cache for a DRAM-based main memory. Neglect the possibility of other caches or levels of the memory hierarchy below main memory. If a cache is improved, increasing the typical hit rate from 98% to 99%, which one of the following would best characterize the likely decrease in typical access time?
(a) 0% (b) 1% (c) 10% (d) 100%
Page 3 of 15

Problem 2. (8 points):
Integer encoding. Fill in the blanks in the table below with the number described in the first column of each row. You can give your answers as unexpanded simple arithmetic expressions (such as 15213 + 42); you should not have trouble fitting your answers into the space provided.
For this problem, assume a 6-bit word size. Description
(unsigned) ((int) 4) (unsigned) ((int) -7) (((unsigned) 0x21) << 1) & 0x3F) (int) (20 + 12) (! 0x15) > 16
Page 4 of 15

Problem 3. (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 3 9 3/16 15/2
Floating Point Bits
Rounded value 1/4
Page 5 of 15

Problem 4. (4 points):
Functions. I never learned to properly comment my code, and now I’ve forgotten what this function does. Help me out by looking at the assembly code and reconstructing the C code for this recursive function. Fill in the blanks:
unsigned mystery1(unsigned n) {
if(________)
return 1 + mystery1(________);
pushl %ebp
movl %esp, %ebp
subl \$8, %esp
cmpl \$0, 8(%ebp)
movl \$1, -4(%ebp)
jmp .L3 .L2:
movl 8(%ebp), %eax
movl %eax, (%esp)
call mystery1
movl %eax, -4(%ebp)
movl -4(%ebp), %eax
Page 6 of 15

Problem 5. (10 points):
Stack discipline. Consider the following C code and assembly code for a recursive function:
int gcd(int a, int b)
return a; }
return gcd(b, a % b);
push %ebp
mov %esp,%ebp
sub \$0x10,%esp
mov 0x8(%ebp),%eax
mov 0xc(%ebp),%ecx
test %ecx,%ecx
je 0x80483b7
mov %eax,%edx
sar \$0x1f,%edx
idiv %ecx
mov %edx,0x4(%esp)
mov %ecx,(%esp)
call 0x8048394
0x08048394 <+0>:
0x08048395 <+1>:
0x08048397 <+3>:
0x0804839a <+6>:
0x0804839d <+9>:
0x080483a0 <+12>:
0x080483a2 <+14>:
0x080483a4 <+16>:
0x080483a6 <+18>:
0x080483a9 <+21>:
0x080483ab <+23>:
0x080483af <+27>:
0x080483b2 <+30>:
0x080483b7 <+35>: leave
0x080483b8 <+36>: ret
Imagine that a program makes the procedure call gcd(213, 18). Also imagine that prior to the invoca- tion, the value of %esp is 0xffff1000—that is, 0xffff1000 is the value of %esp immediately before the execution of the call instruction.
A. Note that the call gcd(213, 18) will result in the following function invocations: gcd(213, 18), gcd(18, 15), gcd(15, 3), and gcd(3, 0). Using the provided code and your knowl- edge of IA32 stack discipline, fill in the stack diagram with the values that would be present imme- diately before the execution of the leave instruction for gcd(15, 3). Supply numerical values wherever possible, and cross out each blank for which there is insufficient information to complete with a numerical value.
Hints: The following set of C style statements describes an approximation of the operation of the instructionidiv %ecx,where‘/’isthedivisionoperatorand‘%’isthemodulooperator:
%eax = %eax / %ecx
%edx = %eax % %ecx
Also,recallthatleaveisequivalenttomovl %ebp, %esp; popl %ebp
Page 7 of 15

+——————————–+
| | 0xffff1008
+——————————–+
| | 0xffff1004
+——————————–+
| | 0xffff1000
+——————————–+
| | 0xffff0ffc
+——————————–+
| | 0xffff0ff8
+——————————–+
| | 0xffff0ff4
+——————————–+
| | 0xffff0ff0
+——————————–+
| | 0xffff0fec
+——————————–+
| | 0xffff0fe8
+——————————–+
| | 0xffff0fe4
+——————————–+
| | 0xffff0fe0
+——————————–+
| | 0xffff0fdc
+——————————–+
| | 0xffff0fd8
+——————————–+
| | 0xffff0fd4
+——————————–+
| | 0xffff0fd0
+——————————–+
| | 0xffff0fcc
+——————————–+
| | 0xffff0fc8
+——————————–+
| | 0xffff0fc4
+——————————–+
| | 0xffff0fc0
+——————————–+
| | 0xffff0fbc
+——————————–+
| | 0xffff0fb8
+——————————–+
| | 0xffff0fb4
+——————————–+
| | 0xffff0fb0
+——————————–+
B. What are the values of %esp and %ebp immediately before the execution of the ret instruction for gcd(15, 3)?
Page 8 of 15

Problem 6. (8 points):
Structs. Consider the following struct on an x86-64 Linux machine:
struct my_struct {
long long b;
float *d;
unsigned char e;
A. (4 points)
Please lay out the struct in memory below. Shade in any bytes used for padding and clearly indicate
the end of the struct.
+—-+—-+—-+—-+—-+—-+—-+—-+ 0x0 | | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+ 0x8 | | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+ 0x10| | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+ 0x18| | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+ 0x20| | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+ 0x28| | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+ 0x30| | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+ 0x38| | | | | | | | | +—-+—-+—-+—-+—-+—-+—-+—-+
Page 9 of 15

B. (3 points)
Consider the following C function:
void foo(struct my_struct *st) {
st->a = ’e’;
st->d = NULL;
st->c = 0x213;
printf(“%lld %p %hhu
”, st->b, &st->f, st->e);
Fill in the missing pieces of the diassembled version of foo.
(gdb) disassemble foo
Dump of assembler code for function foo:
0x00000000004004e4 <+0>:
0x00000000004004e8 <+4>:
0x00000000004004eb <+7>:
0x00000000004004f3 <+15>:
0x00000000004004f9 <+21>:
0x00000000004004fd <+25>:
0x0000000000400501 <+29>:
0x0000000000400505 <+33>:
0x000000000040050a <+38>:
0x000000000040050f <+43>:
0x0000000000400518 <+52>: retq
sub \$0x8,%rsp
movb \$0x65,________(%rdi)
movq \$0x0,________(%rdi)
movw \$0x213,________(%rdi)
movzbl ________(%rdi),%ecx
lea ________(%rdi),%rdx
mov ________(%rdi),%rsi
mov \$0x40062c,%edi
mov \$0x0,%eax
callq 0x4003e0
End of assembler dump.
C. (1 point)
How many bytes is the smallest possible struct containing the same elements as my struct?
(d) None of the above
Page 10 of 15

Problem 7. (10 points):
Switch statements. The problem concerns code generated by GCC for a function involving a switch state- ment. The code uses a jump to index into the jump table:
0x4004b7: jmpq *0x400600(,%rax,8)
Using GDB, we extract the 8-entry jump table:
0x400600: 0x00000000004004d1 0x00000000004004c8
0x400610: 0x00000000004004c8 0x00000000004004be
0x400620: 0x00000000004004c1 0x00000000004004d7
0x400630: 0x00000000004004c8 0x00000000004004be
Here is the block of disassembled code implementing the switch statement:
# on entry: %rdi = x, %rsi = y, %rdx = z
0x4004c0: retq
0x4004d0: retq
0x4004d6: retq
0x4004dc: retq
cmp \$0x7,%edx
ja 0x4004c8
mov %edx,%eax
jmpq *0x400600(,%rax,8)
mov %edi,%eax
mov \$0x3,%eax
jmp 0x4004da
mov %esi,%eax
nopw 0x(%rax,%rax,1)
mov %edi,%eax
and \$0x19,%eax
lea (%rdi,%rdi,1),%eax
Page 11 of 15

Fill in the blank portions of the C code below to reproduce the function corresponding to this object code.
int test(int x, int y, int z)
int result = 3;
_____________;
result = _____________;
result = _____________;
result = _____________;
result = _____________;
return result;
Page 12 of 15

Problem 8. (4 points):
Cache operation. Assume a cache memory with the following properties:
• The cache size (C) is 512 bytes (contains 512 data bytes)
• The cache uses an LRU (least-recently used) policy for eviction. • The cache is initially empty.
Suppose that for the following sequence of addresses sent to the cache, 0, 2, 4, 8, 16, 32, the hit rate is 0.33. Then what is the block size (B) of the cache?
A. B=4bytes
B. B=8bytes
C. B = 16 bytes
D. None of the above.
Page 13 of 15

Problem 9. (6 points):
Miss rate analysis. Listed below are two matrix multiply functions. The first, matrix multiply com- putes C = AB, a standard matrix multiply. The second, matrix multiply t, computes C = ABT , A times the transpose of B.
void matrix_multiply(float A[N][N], float B[N][N], float C[N][N])
/* Computes C = A*B. Assumes C starts as all zeros. */
int i,j,k;
for (i=0; iCS代考 加微信: cscodehelp QQ: 2235208643 Email: kyit630461@163.com