Bomb Lab问题,求分析代码,汇编看不懂。 10C

Dump of assembler code for function phase_2:
=> 0x0000000000401107 <+0>: mov %rbx,-0x18(%rsp)
0x000000000040110c <+5>: mov %rbp,-0x10(%rsp)
0x0000000000401111 <+10>: mov %r12,-0x8(%rsp)
0x0000000000401116 <+15>: sub $0x38,%rsp
0x000000000040111a <+19>: mov %rsp,%rsi
0x000000000040111d <+22>: callq 0x401294
0x0000000000401122 <+27>: mov %rsp,%rbx
0x0000000000401125 <+30>: lea 0xc(%rsp),%r12
0x000000000040112a <+35>: mov $0x0,%ebp
0x000000000040112f <+40>: mov (%rbx),%eax
0x0000000000401131 <+42>: cmp 0xc(%rbx),%eax
0x0000000000401134 <+45>: je 0x40113b
0x0000000000401136 <+47>: callq 0x401272
0x000000000040113b <+52>: add (%rbx),%ebp
0x000000000040113d <+54>: add $0x4,%rbx
0x0000000000401141 <+58>: cmp %r12,%rbx
0x0000000000401144 <+61>: jne 0x40112f
0x0000000000401146 <+63>: test %ebp,%ebp
0x0000000000401148 <+65>: jne 0x40114f
0x000000000040114a <+67>: callq 0x401272
0x000000000040114f <+72>: mov 0x20(%rsp),%rbx
0x0000000000401154 <+77>: mov 0x28(%rsp),%rbp
---Type to continue, or q to quit---return
0x0000000000401159 <+82>: mov 0x30(%rsp),%r12
0x000000000040115e <+87>: add $0x38,%rsp
0x0000000000401162 <+91>: retq

2个回答

CSAPP有两章专门讲汇编语法的,书看明白这个实验才能做。

你这么宽泛的问题,怎么回答?

汇编工具masm.exe

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