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100000f70: 100000f71: pushq %rbp 55 48 89 e5 89 7d fc c7 45 18 02 ee ee ee 8b 7d f8 movq movl Krsp, Krbp Kedi, -4(%rbp) int foolint num) { int val = 2; int result = val - num; 10000er81: 100000f74: 100000177: $2, -8(%rbp movl -8(%rbp), %edi -4 (Krbp), Nedi %edi, -12(Nrbp) -12 (%rbp), %eax 10000017e: movl Of af 7d fc imull movl 100000f85: 89 7d 14 return result; 100000f88: 8b 45 f4 movl 100000f8b: 5d %rbp popq retq 100000f8c: c3 int foo(int num) : pushg &rbp movg trap, rbp movl tedi, -4 (%rbp)e

100000f70: 100000f71: pushq %rbp 55 48 89 e5 89 7d fc c7 45 18 02 ee ee ee 8b 7d f8 movq movl Krsp, Krbp Kedi, -4(%rbp) int foolint num) { int val = 2; int result = val - num; 10000er81: 100000f74: 100000177: $2, -8(%rbp movl -8(%rbp), %edi -4 (Krbp), Nedi %edi, -12(Nrbp) -12 (%rbp), %eax 10000017e: movl Of af 7d fc imull movl 100000f85: 89 7d 14 return result; 100000f88: 8b 45 f4 movl 100000f8b: 5d %rbp popq retq 100000f8c: c3 int foo(int num) : pushg &rbp movg trap, rbp movl tedi, -4 (%rbp)e

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. Given the following code snippet and assembly instructions, align the code with the corresponding assembly. The first line has already been done for you 23fooii: 10000er68: 55 pusha Krbp 10000ef61: 48 89 e5 Krsp, Krbp 89 7d fc mova movl 10000ef64: %edi, -4(%rbp) %esi, -8(%rbp) movl $3, -4(%rbp), %esi %esi, -16(%rbp) -8(Krbp), %esi -12(Xrbp), Nesi Kesi, -20(%rbp) -12(Xrbp), Xesi -16(Nrbp), %esi %esi, -24 (%rbp) -24 (%rbp), %eax 10000ef67: movl c7 45 14 02 ee ee ee imull movl movl 89 75 18 int foo(int num1, int num2){ int val1 = 2; int val2 = 3 * num1; int val3 = num2 + val1; int result = val1 - val2; 1e000e17e: 10000ef6a: $2, -12(Nrbp) 100000f71: 6b 75 fc 03 10000ef75: 18000ef78: 89 75 fe 8b 75 f8 03 75 14 addl movl movl 10000ef7b: 89 75 ec 8b 75 f4 2b 75 fe 10000er81: 10000ef84: subl return result; 89 75 e8 8b 45 e8 5d movl movl 10000er87: 10000ef8a: 10000ef8d: } Krbp popa 1000eef8e: c3 reta int foo (int numl, int num2) : pushg &rbp movg trsp, rbp movl tedi, movl sesi, -8 (8rbp) e -4 (rbp)e www w
Transcribed Image Text:2. Given the following code snippet and assembly instructions, align the code with the corresponding assembly. The first line has already been done for you 23fooii: 10000er68: 55 pusha Krbp 10000ef61: 48 89 e5 Krsp, Krbp 89 7d fc mova movl 10000ef64: %edi, -4(%rbp) %esi, -8(%rbp) movl $3, -4(%rbp), %esi %esi, -16(%rbp) -8(Krbp), %esi -12(Xrbp), Nesi Kesi, -20(%rbp) -12(Xrbp), Xesi -16(Nrbp), %esi %esi, -24 (%rbp) -24 (%rbp), %eax 10000ef67: movl c7 45 14 02 ee ee ee imull movl movl 89 75 18 int foo(int num1, int num2){ int val1 = 2; int val2 = 3 * num1; int val3 = num2 + val1; int result = val1 - val2; 1e000e17e: 10000ef6a: $2, -12(Nrbp) 100000f71: 6b 75 fc 03 10000ef75: 18000ef78: 89 75 fe 8b 75 f8 03 75 14 addl movl movl 10000ef7b: 89 75 ec 8b 75 f4 2b 75 fe 10000er81: 10000ef84: subl return result; 89 75 e8 8b 45 e8 5d movl movl 10000er87: 10000ef8a: 10000ef8d: } Krbp popa 1000eef8e: c3 reta int foo (int numl, int num2) : pushg &rbp movg trsp, rbp movl tedi, movl sesi, -8 (8rbp) e -4 (rbp)e www w
1. Given the following code snippet and assembly instructions, align the code with the corresponding assembly. The first line has already been done for you. -_23fooi: 100000f70: 55 pushq %rbp %rsp, %rbp %edi, -4(%rbp) movl 100000f71: 48 89 e5 movq int foolint num){ int val = 2; 89 7d fc 100000f74: 100000f77: 100000f7e: int result = val num; 10000e181: movl c7 45 f8 02 ee ee 00 movl imull movl movl $2, -8(%rbp) -8(%rbp), %edi -4(%rbp), %edi %edi, -12(%rbp) -12 (%rbp), %eax 8b 7d f8 ef af 7d fc 100000f85: 100000f88: 89 7d f4 return result; 8b 45 f4 100000f8b: 5d %rbp popa retq 100000f8c: c3 int foo (int num) : Rushg &rbp movg rsp, trbp movl šedi, -4 (8rbp)e
Transcribed Image Text:1. Given the following code snippet and assembly instructions, align the code with the corresponding assembly. The first line has already been done for you. -_23fooi: 100000f70: 55 pushq %rbp %rsp, %rbp %edi, -4(%rbp) movl 100000f71: 48 89 e5 movq int foolint num){ int val = 2; 89 7d fc 100000f74: 100000f77: 100000f7e: int result = val num; 10000e181: movl c7 45 f8 02 ee ee 00 movl imull movl movl $2, -8(%rbp) -8(%rbp), %edi -4(%rbp), %edi %edi, -12(%rbp) -12 (%rbp), %eax 8b 7d f8 ef af 7d fc 100000f85: 100000f88: 89 7d f4 return result; 8b 45 f4 100000f8b: 5d %rbp popa retq 100000f8c: c3 int foo (int num) : Rushg &rbp movg rsp, trbp movl šedi, -4 (8rbp)e
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