我创建了这个简单且毫无意义的汇编 (Y86) 代码,以查看在使用指令调用、pushl、popl 和 ret 时我是否理解堆栈中发生的所有事情。
就像我说的,这段代码毫无意义,它只是用于测试/学习目的。虽然,所有内存地址都是正确(有希望的)计算出来的,而且不是随机的。
汇编代码如下:
| .pos 0
0x00 | irmovl Stack, %esp
0x06 | rrmovl %esp, %ebp
0x08 | irmovl $5, %eax
0x0E | call func
0x13 | halt
0x14 | func:
0x14 | pushl %ebp
0x16 | rrmovl %esp, %ebp
0x18 | pushl %eax
0x1A | popl %eax
0x1C | popl %ebp
0x1E | ret
| .pos 50
0x32 | Stack: .long 0
以下是我最好的绘制堆栈并解释每个步骤(指令)对堆栈的作用。请注意,我使用 SP 和 BP 分别指代 %esp 和 %ebp 因为它们被大量使用并且更易于阅读。
我想知道的是我是否把所有事情都做对了,或者我是否错过了任何事情。请随意复制/粘贴您想要的任何内容,并在您的答案中修复一些步骤。
另请注意,我对此的理解非常重要,我周一有一个考试,我需要为此做好准备,我希望你能给我最好的答案。根据您的回答,我可能(或没有)有一些相关问题,我们将在评论部分处理。
- INSTRUCTION: irmovl Stack, %esp
- INSTRUCTION: rrmovl %esp, %ebp
1) Point %esp (SP) and %ebp (BP) to Stack
| ... |
0x2E |-------|
| |
0x32 |-------| <--- SP & BP
- INSTRUCTION: irmovl $5, %eax
1) Sets %eax = 5
- INSTRUCTION: call func
1) Decrements SP by 4 (0x32 -> 0x2E)
2) Saves return address (0x13) in memory location pointed by SP (0x2E)
3) Jumps to "func" memory address (0x14)
| ... |
0x2A |-------|
| 0x13 |
0x2E |-------| <--- SP
| |
0x32 |-------| <--- BP
- INSTRUCTION: pushl %ebp
1) Decrements SP by 4 (0x2E -> 0x2A)
2) Saves BP value (0x32) in memory location pointed by SP (0x2A)
| ... |
0x26 |-------|
| 0x32 |
0x2A |-------| <--- SP
| 0x13 |
0x2E |-------|
| |
0x32 |-------| <--- BP
- INSTRUCTION: rrmovl %esp, %ebp
1) Sets BP = SP (0x32 -> 0x2A)
| ... |
0x26 |-------|
| 0x32 |
0x2A |-------| <--- SP & BP
| 0x13 |
0x2E |-------|
| |
0x32 |-------|
- INSTRUCTION: pushl %eax
1) Decrements SP by 4 (0x2A -> 0x26)
2) Saves %eax value (5) in memory location pointed by SP (0x26)
| ... |
0x22 |-------|
| 5 |
0x26 |-------| <--- SP
| 0x32 |
0x2A |-------| <--- BP
| 0x13 |
0x2E |-------|
| |
0x32 |-------|
- INSTRUCTION: popl %eax
1) Saves value (5) in memory location pointed by SP (0x26) in %eax
2) Increments SP by 4 (0x26 -> 0x2A)
| ... |
0x22 |-------|
| 5 |
0x26 |-------|
| 0x32 |
0x2A |-------| <--- SP & BP
| 0x13 |
0x2E |-------|
| |
0x32 |-------|
- INSTRUCTION: popl %ebp
1) Saves value (0x32) in memory location pointed by SP (0x2A) in %ebp
2) Increments SP by 4 (0x2A -> 0x2E)
| ... |
0x22 |-------|
| 5 |
0x26 |-------|
| 0x32 |
0x2A |-------|
| 0x13 |
0x2E |-------| <--- SP
| |
0x32 |-------| <--- BP
- INSTRUCTION: ret
1) Jumps to memory address (0x13) in memory location pointed by SP (0x2E)
2) Increments SP by 4 (0x2E -> 0x32)