因为确实没有一个简单的答案,所以最好的方法是使用 gnu 调试器或文档,但无论如何你最终都会使用 gdb。一种方法是用 C 编写程序,将它们反汇编,然后自己看看调用约定是什么。您可以使用堆栈,也可以像 64 位通常和 32 位系统调用一样轻松地使用寄存器来传递这些简单的值。
//testc.c
int func2(int c, int d)
{
return c-d;
}
int func(int a, int b)
{
a+=2;
b++;
func2(a,b);
}
//cfile.c
#include <stdio.h>
#include <stdlib.h>
extern int func(int a,int b);
extern int func2(int c,int d);
int main()
{
int res;
int b = 4;
int c = 3;
res=func(b, c);
}
编译
$ gcc -m32 -g -c testc.c
unroot@flerb:~/stacko$ gcc -m32 -g cfile.o testc.o -o a.out
unroot@flerb:~/stacko$ ./a.out
unroot@flerb:~/stacko$ echo $?
0
在 gdb 中
$ gdb -q a.out
Reading symbols from a.out...done.
(gdb) set listsize 50
(gdb) list 0
1 #include <stdio.h>
2 #include <stdlib.h>
3
4 extern int func(int a,int b);
5 extern int func2(int c,int d);
6 int main()
7 {
8 int res;
9 int b = 4;
10 int c = 3;
11 res=func(b, c);
12 }
(gdb) break main
Breakpoint 1 at 0x57c: file cfile.c, line 9.
(gdb) run
Starting program: /home/unroot/stacko/a.out
Breakpoint 1, main () at cfile.c:9
9 int b = 4;
在 main 中断时查看堆栈的内容
(gdb) x/20x $esp
0xffffd290: 0x00000001 0xffffd354 0xffffd35c 0x56555611
0xffffd2a0: 0xffffd2c0 0x00000000 0x00000000 0xf7e12276
0xffffd2b0: 0x00000001 0xf7fad000 0x00000000 0xf7e12276
0xffffd2c0: 0x00000001 0xffffd354 0xffffd35c 0x00000000
0xffffd2d0: 0x00000000 0x00000000 0xf7fad000 0xf7ffdc04
只是为了在参数被推送之前快速浏览一下内容
(gdb) x/20x $esp-0x10
0xffffd280: 0x00000003 0x56557000 0x00000001 0x56555577
0xffffd290: 0x00000001 0xffffd354 0xffffd35c 0x56555611
0xffffd2a0: 0xffffd2c0 0x00000000 0x00000000 0xf7e12276
0xffffd2b0: 0x00000001 0xf7fad000 0x00000000 0xf7e12276
0xffffd2c0: 0x00000001 0xffffd354 0xffffd35c 0x00000000
(gdb) disas
Dump of assembler code for function main:
0x56555560 <+0>: lea ecx,[esp+0x4]
0x56555564 <+4>: and esp,0xfffffff0
0x56555567 <+7>: push DWORD PTR [ecx-0x4]
0x5655556a <+10>: push ebp
0x5655556b <+11>: mov ebp,esp
0x5655556d <+13>: push ebx
0x5655556e <+14>: push ecx
0x5655556f <+15>: sub esp,0x10
0x56555572 <+18>: call 0x565555af <__x86.get_pc_thunk.ax>
0x56555577 <+23>: add eax,0x1a89
=> 0x5655557c <+28>: mov DWORD PTR [ebp-0xc],0x4
0x56555583 <+35>: mov DWORD PTR [ebp-0x10],0x3
0x5655558a <+42>: sub esp,0x8
0x5655558d <+45>: push DWORD PTR [ebp-0x10]
0x56555590 <+48>: push DWORD PTR [ebp-0xc]
0x56555593 <+51>: mov ebx,eax
0x56555595 <+53>: call 0x565555c8 <func>
0x5655559a <+58>: add esp,0x10
0x5655559d <+61>: mov DWORD PTR [ebp-0x14],eax
0x565555a0 <+64>: mov eax,0x0
0x565555a5 <+69>: lea esp,[ebp-0x8]
0x565555a8 <+72>: pop ecx
0x565555a9 <+73>: pop ebx
0x565555aa <+74>: pop ebp
0x565555ab <+75>: lea esp,[ecx-0x4]
0x565555ae <+78>: ret
End of assembler dump.
(gdb) break *0x56555595
Breakpoint 2 at 0x56555595: file cfile.c, line 11.
(gdb) cont
Continuing.
现在看栈的内容,c=3 被压入,然后 b=4(注意我写这个只是为了方便,压入的值是纯值,与一旦它们在堆栈上就代表它们)
Breakpoint 2, 0x56555595 in main () at cfile.c:11
11 res=func(b, c);
(gdb) x/20x $esp
0xffffd280: 0x00000004 0x00000003 0x00000001 0x56555577
0xffffd290: 0x00000001 0xffffd354 0x00000003 0x00000004
0xffffd2a0: 0xffffd2c0 0x00000000 0x00000000 0xf7e12276
0xffffd2b0: 0x00000001 0xf7fad000 0x00000000 0xf7e12276
0xffffd2c0: 0x00000001 0xffffd354 0xffffd35c 0x00000000
(gdb)
所以 func 调用的参数以相反的顺序推送,推送 3 然后推送 4,很明显,因为堆栈正在向下增长到更小的地址。当被调用函数访问这些参数时,有时会先弹出 4 个,然后使用 esp 弹出 3 个到单独的寄存器,或者如下反汇编所示,被调用函数可以通过来自 ebp 的指针访问它们。
(gdb) disas
Dump of assembler code for function main:
0x56555560 <+0>: lea ecx,[esp+0x4]
0x56555564 <+4>: and esp,0xfffffff0
0x56555567 <+7>: push DWORD PTR [ecx-0x4]
0x5655556a <+10>: push ebp
0x5655556b <+11>: mov ebp,esp
0x5655556d <+13>: push ebx
0x5655556e <+14>: push ecx
0x5655556f <+15>: sub esp,0x10
0x56555572 <+18>: call 0x565555af <__x86.get_pc_thunk.ax>
0x56555577 <+23>: add eax,0x1a89
0x5655557c <+28>: mov DWORD PTR [ebp-0xc],0x4
0x56555583 <+35>: mov DWORD PTR [ebp-0x10],0x3
0x5655558a <+42>: sub esp,0x8
0x5655558d <+45>: push DWORD PTR [ebp-0x10]
0x56555590 <+48>: push DWORD PTR [ebp-0xc]
0x56555593 <+51>: mov ebx,eax
=> 0x56555595 <+53>: call 0x565555c8 <func>
0x5655559a <+58>: add esp,0x10
0x5655559d <+61>: mov DWORD PTR [ebp-0x14],eax
0x565555a0 <+64>: mov eax,0x0
0x565555a5 <+69>: lea esp,[ebp-0x8]
0x565555a8 <+72>: pop ecx
0x565555a9 <+73>: pop ebx
0x565555aa <+74>: pop ebp
0x565555ab <+75>: lea esp,[ecx-0x4]
0x565555ae <+78>: ret
End of assembler dump.
进入函数
(gdb) stepi
func (a=4, b=3) at testc.c:9
9 {
(gdb) disas
Dump of assembler code for function func:
=> 0x565555c8 <+0>: push ebp
0x565555c9 <+1>: mov ebp,esp
0x565555cb <+3>: call 0x565555af <__x86.get_pc_thunk.ax>
0x565555d0 <+8>: add eax,0x1a30
0x565555d5 <+13>: add DWORD PTR [ebp+0x8],0x2
0x565555d9 <+17>: add DWORD PTR [ebp+0xc],0x1
0x565555dd <+21>: push DWORD PTR [ebp+0xc]
0x565555e0 <+24>: push DWORD PTR [ebp+0x8]
0x565555e3 <+27>: call 0x565555b3 <func2>
0x565555e8 <+32>: add esp,0x8
0x565555eb <+35>: nop
0x565555ec <+36>: leave
0x565555ed <+37>: ret
End of assembler dump.
C 调用函数将返回值压入堆栈,没有任何明显暗示它会这样做,但它在那里占用空间
(gdb) x/20x $esp
0xffffd27c: 0x5655559a 0x00000004 0x00000003 0x00000001
0xffffd28c: 0x56555577 0x00000001 0xffffd354 0x00000003
0xffffd29c: 0x00000004 0xffffd2c0 0x00000000 0x00000000
0xffffd2ac: 0xf7e12276 0x00000001 0xf7fad000 0x00000000
0xffffd2bc: 0xf7e12276 0x00000001 0xffffd354 0xffffd35c
然后我们进入 func 的实质,其中从 ebp 的偏移用于将 2 加到 a(第二个压入堆栈的函数,因此更接近堆栈上的 ebp)并将 1 加到 b它被推到它之前
(gdb) step
10 a+=2;
(gdb) print/x $ebp
$1 = 0xffffd278
(gdb) x/20x $ebp
0xffffd278: 0xffffd2a8 0x5655559a 0x00000004 0x00000003
0xffffd288: 0x00000001 0x56555577 0x00000001 0xffffd354
0xffffd298: 0x00000003 0x00000004 0xffffd2c0 0x00000000
0xffffd2a8: 0x00000000 0xf7e12276 0x00000001 0xf7fad000
0xffffd2b8: 0x00000000 0xf7e12276 0x00000001 0xffffd354
(gdb) disas
Dump of assembler code for function func:
0x565555c8 <+0>: push ebp
0x565555c9 <+1>: mov ebp,esp
0x565555cb <+3>: call 0x565555af <__x86.get_pc_thunk.ax>
0x565555d0 <+8>: add eax,0x1a30
=> 0x565555d5 <+13>: add DWORD PTR [ebp+0x8],0x2 //a+=2
0x565555d9 <+17>: add DWORD PTR [ebp+0xc],0x1 //b++
0x565555dd <+21>: push DWORD PTR [ebp+0xc] //push b
0x565555e0 <+24>: push DWORD PTR [ebp+0x8] //push a
0x565555e3 <+27>: call 0x565555b3 <func2>
0x565555e8 <+32>: add esp,0x8
0x565555eb <+35>: nop
0x565555ec <+36>: leave
0x565555ed <+37>: ret
End of assembler dump.
中断调用并再次检查堆栈
(gdb) break *0x565555e3
Breakpoint 3 at 0x565555e3: file testc.c, line 12.
(gdb) cont
Continuing.
Breakpoint 3, 0x565555e3 in func (a=6, b=4) at testc.c:12
12 func2(a,b);
(gdb) disas
Dump of assembler code for function func:
0x565555c8 <+0>: push ebp
0x565555c9 <+1>: mov ebp,esp
0x565555cb <+3>: call 0x565555af <__x86.get_pc_thunk.ax>
0x565555d0 <+8>: add eax,0x1a30
0x565555d5 <+13>: add DWORD PTR [ebp+0x8],0x2
0x565555d9 <+17>: add DWORD PTR [ebp+0xc],0x1
0x565555dd <+21>: push DWORD PTR [ebp+0xc]
0x565555e0 <+24>: push DWORD PTR [ebp+0x8]
=> 0x565555e3 <+27>: call 0x565555b3 <func2>
0x565555e8 <+32>: add esp,0x8
0x565555eb <+35>: nop
0x565555ec <+36>: leave
0x565555ed <+37>: ret
(gdb) x/20x $esp
0xffffd270: 0x00000006 0x00000004 0xffffd2a8 0x5655559a
0xffffd280: 0x00000006 0x00000004 0x00000001 0x56555577
0xffffd290: 0x00000001 0xffffd354 0x00000003 0x00000004
0xffffd2a0: 0xffffd2c0 0x00000000 0x00000000 0xf7e12276
0xffffd2b0: 0x00000001 0xf7fad000 0x00000000 0xf7e12276
Again our variables have been pushed on the stack, again in reverse order, as always, but you could control this too in assembly if you feel like being difficult.
(gdb) stepi
func2 (c=6, d=4) at testc.c:3
3 {
(gdb) disas
Dump of assembler code for function func2:
=> 0x565555b3 <+0>: push ebp
0x565555b4 <+1>: mov ebp,esp
0x565555b6 <+3>: call 0x565555af <__x86.get_pc_thunk.ax>
0x565555bb <+8>: add eax,0x1a45
0x565555c0 <+13>: mov eax,DWORD PTR [ebp+0x8]
0x565555c3 <+16>: sub eax,DWORD PTR [ebp+0xc]
0x565555c6 <+19>: pop ebp
0x565555c7 <+20>: ret
End of assembler dump.
Again the call pushes the return address onto the stack before transfering control to the called function.
(gdb) x/20x $esp
0xffffd26c: 0x565555e8 0x00000006 0x00000004 0xffffd2a8
0xffffd27c: 0x5655559a 0x00000006 0x00000004 0x00000001
0xffffd28c: 0x56555577 0x00000001 0xffffd354 0x00000003
0xffffd29c: 0x00000004 0xffffd2c0 0x00000000 0x00000000
0xffffd2ac: 0xf7e12276 0x00000001 0xf7fad000 0x00000000
在被调用函数的实质上继续和中断。
(gdb) break *0x565555c0
Breakpoint 4 at 0x565555c0: file testc.c, line 4.
(gdb) cont
Continuing.
Breakpoint 4, func2 (c=6, d=4) at testc.c:4
4 return c-d;
(gdb) disas
Dump of assembler code for function func2:
0x565555b3 <+0>: push ebp
0x565555b4 <+1>: mov ebp,esp
0x565555b6 <+3>: call 0x565555af <__x86.get_pc_thunk.ax>
0x565555bb <+8>: add eax,0x1a45
=> 0x565555c0 <+13>: mov eax,DWORD PTR [ebp+0x8]
0x565555c3 <+16>: sub eax,DWORD PTR [ebp+0xc]
0x565555c6 <+19>: pop ebp
0x565555c7 <+20>: ret
End of assembler dump.
在 func2 的实质检查堆栈,访问变量并发生减法:
(gdb) x/20x $esp
0xffffd268: 0xffffd278 0x565555e8 0x00000006 0x00000004
0xffffd278: 0xffffd2a8 0x5655559a 0x00000006 0x00000004
0xffffd288: 0x00000001 0x56555577 0x00000001 0xffffd354
0xffffd298: 0x00000003 0x00000004 0xffffd2c0 0x00000000
0xffffd2a8: 0x00000000 0xf7e12276 0x00000001 0xf7fad000
(gdb) x/20x $ebp
0xffffd268: 0xffffd278 0x565555e8 0x00000006 0x00000004
0xffffd278: 0xffffd2a8 0x5655559a 0x00000006 0x00000004
0xffffd288: 0x00000001 0x56555577 0x00000001 0xffffd354
0xffffd298: 0x00000003 0x00000004 0xffffd2c0 0x00000000
0xffffd2a8: 0x00000000 0xf7e12276 0x00000001 0xf7fad000
所以这里 [ebp+0x8] = 6 和 [ebp+0xc] = 4 并用 eax 寄存器中的减法指令修改值,将结果返回到 eax 寄存器中。
C 的默认约定是让调用者在将控制权转移给被调用者之前推送返回地址,并让被调用者调整堆栈和基指针,但是当你调用自己的函数时,你可以做任何你想做的事情并返回您自己的功能。在这里,我使用 C 程序来说明 C 的作用,但是如果您使用 C 调用一个汇编程序,该程序又调用另一个汇编程序,那么您可以显式控制这些汇编程序之间的基指针(如果需要)。您可以选择手动调整它们并跳转到您的第二个函数调用,这将绕过推送返回地址和设置堆栈的自动调用过程,并不完全有用;或者汇编调用指令将使用相同的过程来初始化被调用的函数,并且您可以期望偏移量与 C 函数中的相同。
Good link on x86 Call conventions including cdecl and stdcall
和
A handy syscall reference showing use of registers for function calls
这是一个使用寄存器传递 32 位变量的简单示例,不需要堆栈。
//testasm32.asm
部分 .text
全局_start
_start:
mov ecx, hello1
call print_string
mov ecx, hello2
call print_string
mov eax, 1
int 0x80
print_string:
mov eax, 4
mov ebx, 1
mov edx, 6
int 0x80
ret
section .data
hello1 db "Hello1"
hello2 db "Hello2"
$ nasm -f elf32 testasm32.asm
unroot@flerb:~/LearningLinuxBinaryAnalysis_Code$ ld -m elf_i386 -o testasm32 testasm32.o
unroot@flerb:~/LearningLinuxBinaryAnalysis_Code$ ./testasm32
Hello1Hello2
$ echo $?
1
有趣的是,ebx 保存了退出系统调用的错误代码,因此它返回 1,因为 ebx 在 print_string 命令中被设置为 1 并且未被清除。