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. code locations Consider the following C code and the corresponding assembly code: oid baz(int a, int * p) pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; c = a-*p; movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) if (c) cmpl $0, -4(%rbp) je L2 { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) movl 24(%rbp),%rax movl (%rax),%eax subl *p -= d; -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: }

. code locations Consider the following C code and the corresponding assembly code: oid baz(int a, int * p) pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; c = a-*p; movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) if (c) cmpl $0, -4(%rbp) je L2 { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) movl 24(%rbp),%rax movl (%rax),%eax subl *p -= d; -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: }

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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2. code locations Consider the following C code and the corresponding assembly code: void baz(int a, int * p) { pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; C = a-*p; movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) cmpl $0, -4(%rbp) je L2 if (c) { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) *p -= d; movl 24(%rbp),%rax movl (%rax),%eax subl -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: (continued on next page)
Transcribed Image Text:2. code locations Consider the following C code and the corresponding assembly code: void baz(int a, int * p) { pushq %rbp movq %rsp,%rbp subq $16,%rsp int c; C = a-*p; movl %ecx,16(%rbp) movq %rdx, 24(%rbp) movq 24(%rbp), %rax movl (%rax),%eax movl 16(%rbp),%edx subl %eax,%edx movl %edx,%eax movl %eax,-4(%rbp) cmpl $0, -4(%rbp) je L2 if (c) { int d = c*2; movl -4(%rbp),%eax addl %eax,%eax movl %eax,-8(%rbp) *p -= d; movl 24(%rbp),%rax movl (%rax),%eax subl -8(%rbp),%eax movl %eax,%edx movq 24(%rbp), %rax movl %edx, (%rax) jmp L4: (continued on next page)
else { int e = *p/2; L2: movq 24(%rbp),%rax movl (%rax),%eax movl %eax, %edx shrl $31, %edx addl %edx, %eax sarl movl %eax,-12(%rbp) %eax movl 24(%rbp),%rax movl (%rax), %edx movl -12(%rbp), %eax subl addl *p += e-a; 16(%rbp),%eax %eax,%edx movq 24(%rbp), %rax movl %edx,(%rax) } } L4: addq $16, %rsp popq %rbp ret Give the memory (not register!) location for each of the 5 variables in this function a C d e
Transcribed Image Text:else { int e = *p/2; L2: movq 24(%rbp),%rax movl (%rax),%eax movl %eax, %edx shrl $31, %edx addl %edx, %eax sarl movl %eax,-12(%rbp) %eax movl 24(%rbp),%rax movl (%rax), %edx movl -12(%rbp), %eax subl addl *p += e-a; 16(%rbp),%eax %eax,%edx movq 24(%rbp), %rax movl %edx,(%rax) } } L4: addq $16, %rsp popq %rbp ret Give the memory (not register!) location for each of the 5 variables in this function a C d e
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