Question

我试图理解由以下C代码生成的汇编指令：

#include<stdio.h>
#include<unistd.h>

int main()
{    
  char *happy[2];
  happy[0] = "/bin/sh";
  happy[1] = NULL;
  execve(happy[0],happy,NULL);    
}

不幸的是我无法理解这个汇编代码任何人都可以向我解释这段代码吗？我可以确定rax包含/bin/sh的地址，但我不知道在这里使用rsi寄存器。

 Dump of assembler code for function main:
   0x0000000000400a0d <+0>:     push   %rbp
   0x0000000000400a0e <+1>:     mov    %rsp,%rbp
   0x0000000000400a11 <+4>:     sub    $0x20,%rsp
   0x0000000000400a15 <+8>:     mov    %fs:0x28,%rax
   0x0000000000400a1e <+17>:    mov    %rax,-0x8(%rbp)
   0x0000000000400a22 <+21>:    xor    %eax,%eax
   0x0000000000400a24 <+23>:    lea    0x8c4d9(%rip),%rax        # 0x48cf04
   0x0000000000400a2b <+30>:    mov    %rax,-0x20(%rbp)
   0x0000000000400a2f <+34>:    movq   $0x0,-0x18(%rbp)
   0x0000000000400a37 <+42>:    mov    -0x20(%rbp),%rax
   0x0000000000400a3b <+46>:    lea    -0x20(%rbp),%rcx
   0x0000000000400a3f <+50>:    mov    $0x0,%edx
   0x0000000000400a44 <+55>:    mov    %rcx,%rsi
   0x0000000000400a47 <+58>:    mov    %rax,%rdi
   0x0000000000400a4a <+61>:    callq  0x432b10 <execve>
   0x0000000000400a4f <+66>:    mov    $0x0,%eax
   0x0000000000400a54 <+71>:    mov    -0x8(%rbp),%rdx
   0x0000000000400a58 <+75>:    xor    %fs:0x28,%rdx
   0x0000000000400a61 <+84>:    je     0x400a68 <main+91>
   0x0000000000400a63 <+86>:    callq  0x435730 <__stack_chk_fail_local>
   0x0000000000400a68 <+91>:    leaveq
   0x0000000000400a69 <+92>:    retq
End of assembler dump.
`

寻求帮助：解释C编译器生成的汇编代码

0 个答案: