Consider the following assembly language code: LCO: string "answer %d\n" text globl main type main, @function main: pushq %rbp movq %rsp, %rbp subq $16, %rsp movl $12, -8(%rbp) movl $3, -4(%rbp) movi $0, -16(%rbp) movl -8(%rbp), %eax -4(%rbp), %eax movi %eax, -12(%rbp) xorl jmp 12 13: movl -12(%rbp), %eax andl $1, %eax addi %eax, -16(%rbp) sarl -12(%rbp) L2: cmpl $0, -12(%rbp) jne 13 movi -16(%rbp), %eax movl %eax, %esi leaq LCO(%rip), %rdi movi $0, %eax printf@PLT movl $0, %eax call leave ret This code came from skeleton C file below after optimizing with 00. This means "gcc -00 -5 fno- asynchronous-unwind-tables" command was used to convert C File into assembly file. (The last option was used to disable cfi directives) Complete the C code given below using the provided assembly code One approach may be to ignore the skeleton file and create an equivalent C code from the assembly code, and then rewrite the code to fit the skeleton file. int main(){ int a - b- 3: int count - for(int e= a b; e!- o; ee count +- (e_ 1); printf("answer %d\n", count); return o;
Consider the following assembly language code: LCO: string "answer %d\n" text globl main type main, @function main: pushq %rbp movq %rsp, %rbp subq $16, %rsp movl $12, -8(%rbp) movl $3, -4(%rbp) movi $0, -16(%rbp) movl -8(%rbp), %eax -4(%rbp), %eax movi %eax, -12(%rbp) xorl jmp 12 13: movl -12(%rbp), %eax andl $1, %eax addi %eax, -16(%rbp) sarl -12(%rbp) L2: cmpl $0, -12(%rbp) jne 13 movi -16(%rbp), %eax movl %eax, %esi leaq LCO(%rip), %rdi movi $0, %eax printf@PLT movl $0, %eax call leave ret This code came from skeleton C file below after optimizing with 00. This means "gcc -00 -5 fno- asynchronous-unwind-tables" command was used to convert C File into assembly file. (The last option was used to disable cfi directives) Complete the C code given below using the provided assembly code One approach may be to ignore the skeleton file and create an equivalent C code from the assembly code, and then rewrite the code to fit the skeleton file. int main(){ int a - b- 3: int count - for(int e= a b; e!- o; ee count +- (e_ 1); printf("answer %d\n", count); return o;
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Question
Transcribed Image Text:Consider the following assembly language code:
LCO:
string "answer %d\n"
text
globi main
type main, @function
main:
pushq %rbp
movq %rsp, %rbp
subq $16, %rsp
movl $12, -8(%rbp)
movl $3, -4(%rbp)
movl $0, -16(%rbp)
movi -8(%rbp), %eax
-4(%rbp), %eax
movi %eax, -12(%rbp)
xorl
jmp 12
L3:
movl -12(%rbp), %eax
andl $1, %eax
addi %eax, -16(%rbp)
-12(%rbp)
sarl
L2:
cmpl $0, -12(%rbp)
jne
13
movi -16(%rbp), %eax
movl %eax, %esi
leaq LCO(%rip), %rdi
movl $0, %eax
call
printf@PLT
movi $0, %eax
leave
ret
This code came from skeleton C file below after optimizing with 00. This means "gcc -00 -5 fno-
asynchronous-unwind-tables" command was used to convert C File into assembly file. (The last option was
used to disable cfi directives)
Complete the C code given below using the provided assembly code
One approach may be to ignore the skeleton file and create an equivalent C code from the assembly
code, and then rewrite the code to fit the skeleton file.
int main(){
int a =
b 3:
int count
for(int e = a
b; e != o; e=e
1){
count + (c
1);
}
printf("answer %d\n", count);
return o;
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