留学生考试辅导 assembly Name: ID#: – cscodehelp代写

Name: ID#:
X.500:
Problem 1 (15 pts):
Nearby is a C function col update()
1 typedef struct{ 2 int cur;
3 int step;
4 } colinfo_t;
5 // | | Byte | Byte | Packed | 6 //|Field|Size|Offset| Bits| 7 // |——-+——+——–+——–| 8 // | cur | 4 | +0 | 0-31 | 9 // | step | 4 | +4 | 32-64 |
with associated data and documentation. Re-implement
this function in x86-64 assembly according to the doc-
umentation given. Follow the same flow provided in the C
implementation. The comments below the colinfo t struct
give information about how it lays out in memory and as a
packed argument.
Indicate which registers correspond to which C vari- 10
ables.
.text
.globl col_update
# YOUR CODE BELOW
col_update:
11 int col_update(colinfo_t *info){
12 // Updates current value and step in
13 // colinfo_t pointed by param info. If
14 // infor->cur is invalid, makes no changes 15 // and returns 1 to indicate an
16 // error. Otherwise performs odd or even 17 // update on cur and increments step
18 // returning 0 for success.
19
20 int cur = info->cur;
21 int step = info->step;
22 if(cur <= 0){ 23 return 1; 24 } 25 step++; 26 if(cur % 2 == 1){ 27 cur = cur*3+1; 28 } 29 else{ 30 cur = cur / 2; 31 } 32 info->cur = cur;
33 info->step = step;
34 return 0;
35 }
CS 2021: Practice Exam 2
Fall 2021 University of Minnesota
Exam period: 20 minutes Points available: 40
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WRITE ON EVERY PAGE – Name:
Problem 2 (15 pts): Below is an initial register/memory configuration along with snippets of assembly code. Each snippet is followed by a blank register/memory configuration which should be filled in with the values to reflect changes made by the preceding assembly. The code is continuous so that POS A is followed by POS B.
INITIAL |——-+——-| | REG | Value | |——-+——-| |rax|10| |rdi|20| |rsi|30| | rsp | #3032 | |——-+——-| | MEM | Value | |——-+——-| | #3032 | 250 | |#3028| 1| |#3024| 2| |#3020| 3| |——-+——-|
addl (%rsp), %eax #POSA |——-+——-| | REG | Value | |——-+——-| |rax|| |rdi|| |rsi|| |rsp|| |——-+——-| | MEM | Value | |——-+——-|
| #3032 |
| #3028 |
| #3024 |
| #3020 |
|——-+——-|
addl (%rdi), %rax # POS B |——-+——-| | REG | Value | |——-+——-| |rax| | |rdi| | |rsi| | |rsp| | |——-+——-| | MEM | Value | |——-+——-| | #3032 | | | #3028 | | | #3024 | | | #3020 | | |——-+——-|
Problem 3 (10 pts): Rover Witer is writing an as- sembly function called compval which he will use in C programs. He writes a short C main() function to test compval but is shocked by the results which seem to defy the C and assembly code. Valgrind provides no insight for him. Identify why Rover’s code is behaving so strangely and fix compval so it behaves correctly.
Sample Compile / Run:
> gcc compval_main.c compval_asm.s
> a.out
expect: 0
actual: 19
expect: 0
actual: 50
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19 20 21 22
1 2 3 4 5 6 7 8
// compval_main.c
#include
void compval(int x, int y, int *val);
// compute something based on x and y
// store result at int pointed to by val
int main(){
int expect, actual;
expect = 7 * 2 + 5;
compval(7, 2, &actual); // actual result
printf(“expect: %d
”,expect);
printf(“actual: %d
”,actual);
expect = 5 * 9 + 5; // expected value
compval(5, 9, &actual); // actual result
printf(“expect: %d
”,expect);
printf(“actual: %d
”,actual);
return 0; }
addl %edi, %esi
subq $8, %rsp movq $1, %rdi
movl $100, 4(%rsp) addl %esi, (%rsp,%rdi,4)
movl $300, 0(%rsp) leaq 8(%rsp), %rdi
| | | |
#
.text
.global compval
compval:
imulq %rdi,%rsi
addq $5,%rsi
movq %rsi,(%rdx)
ret
compval_asm.s
// expected value
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