《coredump问题原理探究》Linux x86版4.2节函数的逆向之顺序结构
版权声明：本文为博主原创文章，未经博主允许不得转载。 https://cloud.tencent.com/developer/article/1344435
先看一下例子：
#include <stdio.h>
int main()
{
int a = 0, b = 0, c = 0, d = 0;
scanf( "%d,%d", &a,&b );
a += b;
scanf( "%d", &c );
a += c;
scanf( "%d", &d );
a += d;
return a;
}看一下main函数的汇编：
(gdb) disassemble main
Dump of assembler code for function main:
0x08048570 <+0>: push %ebp
0x08048571 <+1>: mov %esp,%ebp
0x08048573 <+3>: and $0xfffffff0,%esp
0x08048576 <+6>: sub $0x20,%esp
0x08048579 <+9>: movl $0x0,0x1c(%esp)
0x08048581 <+17>: movl $0x0,0x18(%esp)
0x08048589 <+25>: movl $0x0,0x14(%esp)
0x08048591 <+33>: movl $0x0,0x10(%esp)
0x08048599 <+41>: lea 0x18(%esp),%eax
0x0804859d <+45>: mov %eax,0x8(%esp)
0x080485a1 <+49>: lea 0x1c(%esp),%eax
0x080485a5 <+53>: mov %eax,0x4(%esp)
0x080485a9 <+57>: movl $0x80486b4,(%esp)
0x080485b0 <+64>: call 0x8048440 <scanf@plt>
0x080485b5 <+69>: mov 0x1c(%esp),%edx
0x080485b9 <+73>: mov 0x18(%esp),%eax
0x080485bd <+77>: add %edx,%eax
0x080485bf <+79>: mov %eax,0x1c(%esp)
0x080485c3 <+83>: lea 0x14(%esp),%eax
0x080485c7 <+87>: mov %eax,0x4(%esp)
0x080485cb <+91>: movl $0x80486ba,(%esp)
0x080485d2 <+98>: call 0x8048440 <scanf@plt>
0x080485d7 <+103>: mov 0x1c(%esp),%edx
0x080485db <+107>: mov 0x14(%esp),%eax
0x080485df <+111>: add %edx,%eax
0x080485e1 <+113>: mov %eax,0x1c(%esp)
0x080485e5 <+117>: lea 0x10(%esp),%eax
0x080485e9 <+121>: mov %eax,0x4(%esp)
0x080485ed <+125>: movl $0x80486ba,(%esp)
0x080485f4 <+132>: call 0x8048440 <scanf@plt>
0x080485f9 <+137>: mov 0x1c(%esp),%edx
0x080485fd <+141>: mov 0x10(%esp),%eax
0x08048601 <+145>: add %edx,%eax
0x08048603 <+147>: mov %eax,0x1c(%esp)
0x08048607 <+151>: mov 0x1c(%esp),%eax
0x0804860b <+155>: jmp 0x8048615 <main+165>
0x0804860d <+157>: mov %eax,(%esp)
0x08048610 <+160>: call 0x8048460 <_Unwind_Resume@plt>
0x08048615 <+165>: leave
0x08048616 <+166>: ret
End of assembler dump.区区十来行代码，就变成了非常多的汇编语句，非常令人害怕。实际上，不需要那么害怕。
先看一下call指令的地方，由于call指令是调用函数的，所以，用它可以大致定一下这样的范围。
先看一下三个call指令的地址：
0x080485b0 <+64>: call 0x8048440 <scanf@plt>
0x080485d2 <+98>: call 0x8048440 <scanf@plt>
0x080485f4 <+132>: call 0x8048440 <scanf@plt>可以知道，
0x08048570 <+0>: push %ebp
0x08048571 <+1>: mov %esp,%ebp
0x08048573 <+3>: and $0xfffffff0,%esp
0x08048576 <+6>: sub $0x20,%esp
0x08048579 <+9>: movl $0x0,0x1c(%esp)
0x08048581 <+17>: movl $0x0,0x18(%esp)
0x08048589 <+25>: movl $0x0,0x14(%esp)
0x08048591 <+33>: movl $0x0,0x10(%esp)
0x08048599 <+41>: lea 0x18(%esp),%eax
0x0804859d <+45>: mov %eax,0x8(%esp)
0x080485a1 <+49>: lea 0x1c(%esp),%eax
0x080485a5 <+53>: mov %eax,0x4(%esp)
0x080485a9 <+57>: movl $0x80486b4,(%esp)这一段汇编的地址少于第一个call指令地址0x080485b0,所以，它们大概对应于代码：
int a = 0, b = 0, c = 0, d = 0;而
0x080485b5 <+69>: mov 0x1c(%esp),%edx
0x080485b9 <+73>: mov 0x18(%esp),%eax
0x080485bd <+77>: add %edx,%eax
0x080485bf <+79>: mov %eax,0x1c(%esp)
0x080485c3 <+83>: lea 0x14(%esp),%eax
0x080485c7 <+87>: mov %eax,0x4(%esp)
0x080485cb <+91>: movl $0x80486ba,(%esp)这一段汇编由于地址大于0x080485b0，小于0x080485d2，所以，它们大致对应于代码：
a += b;再
0x080485d7 <+103>: mov 0x1c(%esp),%edx
0x080485db <+107>: mov 0x14(%esp),%eax
0x080485df <+111>: add %edx,%eax
0x080485e1 <+113>: mov %eax,0x1c(%esp)
0x080485e5 <+117>: lea 0x10(%esp),%eax
0x080485e9 <+121>: mov %eax,0x4(%esp)
0x080485ed <+125>: movl $0x80486ba,(%esp)这一段汇编地址大于0x080485d2，小于0x080485f4，所以，它们大致对应于代码：
a += c;而
0x080485f9 <+137>: mov 0x1c(%esp),%edx
0x080485fd <+141>: mov 0x10(%esp),%eax
0x08048601 <+145>: add %edx,%eax
0x08048603 <+147>: mov %eax,0x1c(%esp)
0x08048607 <+151>: mov 0x1c(%esp),%eax
0x0804860b <+155>: jmp 0x8048615 <main+165>
0x0804860d <+157>: mov %eax,(%esp)
0x08048610 <+160>: call 0x8048460 <_Unwind_Resume@plt>
0x08048615 <+165>: leave
0x08048616 <+166>: ret 这一段汇编地址大于0x080485f4，所以，它们对应于代码：
a += d;
return a;看，分析汇编，并不是那么让人恐怖。但上面由于有一些指令是编译器生成的，有一些是函数调用时把参数入栈的指令，所以，要筛选出这些指令，仅以第一段汇编为例（即第一个scanf调用前的汇编）：
0x08048570 <+0>: push %ebp
0x08048571 <+1>: mov %esp,%ebp
0x08048573 <+3>: and $0xfffffff0,%esp
0x08048576 <+6>: sub $0x20,%esp
0x08048579 <+9>: movl $0x0,0x1c(%esp)
0x08048581 <+17>: movl $0x0,0x18(%esp)
0x08048589 <+25>: movl $0x0,0x14(%esp)
0x08048591 <+33>: movl $0x0,0x10(%esp)
0x08048599 <+41>: lea 0x18(%esp),%eax
0x0804859d <+45>: mov %eax,0x8(%esp)
0x080485a1 <+49>: lea 0x1c(%esp),%eax
0x080485a5 <+53>: mov %eax,0x4(%esp)
0x080485a9 <+57>: movl $0x80486b4,(%esp)由第三章可知，
0x08048570 <+0>: push %ebp
0x08048571 <+1>: mov %esp,%ebp是属于函数开头的特征指令，所以，这是由编译器自动生成的。
而
0x08048573 <+3>: and $0xfffffff0,%esp
0x08048576 <+6>: sub $0x20,%esp则是用来调整esp和分配局部变量空间的，也是编译器自动生成的。所以，第一段汇编只剩下
0x08048579 <+9>: movl $0x0,0x1c(%esp)
0x08048581 <+17>: movl $0x0,0x18(%esp)
0x08048589 <+25>: movl $0x0,0x14(%esp)
0x08048591 <+33>: movl $0x0,0x10(%esp)
0x08048599 <+41>: lea 0x18(%esp),%eax
0x0804859d <+45>: mov %eax,0x8(%esp)
0x080485a1 <+49>: lea 0x1c(%esp),%eax
0x080485a5 <+53>: mov %eax,0x4(%esp)
0x080485a9 <+57>: movl $0x80486b4,(%esp)是和
int a = 0, b = 0, c = 0, d = 0;对应。由于
scanf( "%d,%d", &a,&b );是有三个参数，根据第三章内容，
0x08048599 <+41>: lea 0x18(%esp),%eax
0x0804859d <+45>: mov %eax,0x8(%esp)
0x080485a1 <+49>: lea 0x1c(%esp),%eax
0x080485a5 <+53>: mov %eax,0x4(%esp)
0x080485a9 <+57>: movl $0x80486b4,(%esp)刚好是把scanf的三个参数入栈的操作，可以通过运行时得到0x80486b4指向的内容：
gdb) tbreak *0x080485b0
Temporary breakpoint 1 at 0x80485b0
(gdb) r
Starting program: /home/buckxu/work/4/1/xuzhina_dump_c4_s1
Temporary breakpoint 1, 0x080485b0 in main ()
(gdb) x /s 0x80486b4
0x80486b4 <__dso_handle+4>: "%d,%d"也就是说，只有
0x08048579 <+9>: movl $0x0,0x1c(%esp)
0x08048581 <+17>: movl $0x0,0x18(%esp)
0x08048589 <+25>: movl $0x0,0x14(%esp)
0x08048591 <+33>: movl $0x0,0x10(%esp)才是和
int a = 0, b = 0, c = 0, d = 0;对应的。
小结：
由于顺序结构的逆向非常考验汇编基础，但如果是有函数调用的话，先找call指令，根据call指令来划分范围，筛选出编译器自动生成的指令。
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