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.LCO .string "ans %d\n" main: .LFBO: pushq %rbp movq %rsp, %rbp subq $16, %rsp movabsq $83162457113523645, %rax movq %rax, -8(%rbp) movl $0, -12(%rbp) jmp .L2 .L3: movq -8(%rbp), %rbx movl %ebx, %ecx andl $1, %ecx movl -12(%rbp), %ebx addl %ecx, %ebx movl %ebx, -12(%rbp) sarq -8(%rbp) .L2: cmpq $0, -8(%rbp) jg .L3 movl -12(%rbp), %ebx movl %ebx, %esi movl $.LCO, %edi movl $0, %ebx call printf leave ret This code came from the following skeleton C file, and optimized with 00, so gcc -00 -S was the exact command to compile the file. Complete the below given C code using the above information. Step 1 might be to ignore the skeleton file and create a C file from the assembly code, and then rewrite the code to fit the skeleton file. #include int main() { long int val= int result= for(;val>. ;val= result += printf("ans %d\n",result); }

.LCO .string "ans %d\n" main: .LFBO: pushq %rbp movq %rsp, %rbp subq $16, %rsp movabsq $83162457113523645, %rax movq %rax, -8(%rbp) movl $0, -12(%rbp) jmp .L2 .L3: movq -8(%rbp), %rbx movl %ebx, %ecx andl $1, %ecx movl -12(%rbp), %ebx addl %ecx, %ebx movl %ebx, -12(%rbp) sarq -8(%rbp) .L2: cmpq $0, -8(%rbp) jg .L3 movl -12(%rbp), %ebx movl %ebx, %esi movl $.LCO, %edi movl $0, %ebx call printf leave ret This code came from the following skeleton C file, and optimized with 00, so gcc -00 -S was the exact command to compile the file. Complete the below given C code using the above information. Step 1 might be to ignore the skeleton file and create a C file from the assembly code, and then rewrite the code to fit the skeleton file. #include int main() { long int val= int result= for(;val>. ;val= result += printf("ans %d\n",result); }

Computer Networking: A Top-Down Approach (7th Edition)
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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Problem 1: Consider the following assembly code: .LCO .string "ans %d\n" main: .LFBO: pushq %rbp movq %rsp, %rbp subq $16, %rsp movabsq $83162457113523645, %rax movq %rax, -8(%rbp) movl $0, -12(%rbp) jmp .L2 .L3: movq -8(%rbp), %rbx movl %ebx, %ecx andl $1, %ecx movl -12(%rbp), %ebx addl %ecx, %ebx movl %ebx, -12(%rbp) sarq -8(%rbp) .L2: cmpq $0, -8(%rbp) jg .L3 movl -12(%rbp), %ebx movl %ebx, %esi movl $.LCO, %edi movl $0, %ebx call printf leave ret This code came from the following skeleton C file, and optimized with 00, so gcc -00 -S was the exact command to compile the file. Complete the below given C code using the above information. Step 1 might be to ignore the skeleton file and create a C file from the assembly code, and then rewrite the code to fit the skeleton file. #include <stdio.h> int main() { long int val= int result= for(;val>. val=. result += printf("ans %d\n",result); }
Transcribed Image Text:Problem 1: Consider the following assembly code: .LCO .string "ans %d\n" main: .LFBO: pushq %rbp movq %rsp, %rbp subq $16, %rsp movabsq $83162457113523645, %rax movq %rax, -8(%rbp) movl $0, -12(%rbp) jmp .L2 .L3: movq -8(%rbp), %rbx movl %ebx, %ecx andl $1, %ecx movl -12(%rbp), %ebx addl %ecx, %ebx movl %ebx, -12(%rbp) sarq -8(%rbp) .L2: cmpq $0, -8(%rbp) jg .L3 movl -12(%rbp), %ebx movl %ebx, %esi movl $.LCO, %edi movl $0, %ebx call printf leave ret This code came from the following skeleton C file, and optimized with 00, so gcc -00 -S was the exact command to compile the file. Complete the below given C code using the above information. Step 1 might be to ignore the skeleton file and create a C file from the assembly code, and then rewrite the code to fit the skeleton file. #include <stdio.h> int main() { long int val= int result= for(;val>. val=. result += printf("ans %d\n",result); }
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