汇编:C编程中变量的内存地址

Assembly: Memory address of variables in C Programming

这是我的 C 代码

C:\Codes>gdb test -q
Reading symbols from C:\Codes\test.exe...done.
(gdb) list 1,15
1       #include<stdio.h>
2
3       int main()
4       {
5               int a = 12345;
6               int b = 0x12345;
7               printf("+-----+-------+---------+----------+\n");
8               printf("| Var | Dec   | Hex     | Address  |\n");
9               printf("+-----+-------+---------+----------+\n");
10              printf("|  a  | %d | 0x%x  | %p |\n",a,a,&a);
11              printf("|  b  | %d | 0x%x | %p |\n",b,b,&b);
12              printf("+-----+-------+---------+----------+\n");
13
14              return 0;
15      }
(gdb) set disassembly-flavor intel

这是标准输出

C:\Codes>test
+-----+-------+---------+----------+
| Var | Dec   | Hex     | Address  |
+-----+-------+---------+----------+
|  a  | 12345 | 0x3039  | 0022FF4C |
|  b  | 74565 | 0x12345 | 0022FF48 |
+-----+-------+---------+----------+

这就是我在 GDB 中看到的内容

(gdb) break 7
Breakpoint 1 at 0x40135e: file test.c, line 7.
(gdb) run
Starting program: C:\Codes/test.exe
[New Thread 4044.0xab0]

Breakpoint 1, main () at test.c:7
7               printf("+-----+-------+---------+----------+\n");
(gdb) disassemble
Dump of assembler code for function main:
   0x00401340 <+0>:     push   ebp
   0x00401341 <+1>:     mov    ebp,esp
   0x00401343 <+3>:     and    esp,0xfffffff0
   0x00401346 <+6>:     sub    esp,0x20
   0x00401349 <+9>:     call   0x401990 <__main>
   0x0040134e <+14>:    mov    DWORD PTR [esp+0x1c],0x3039
   0x00401356 <+22>:    mov    DWORD PTR [esp+0x18],0x12345
=> 0x0040135e <+30>:    mov    DWORD PTR [esp],0x403024
   0x00401365 <+37>:    call   0x401c00 <puts>
   0x0040136a <+42>:    mov    DWORD PTR [esp],0x40304c
   0x00401371 <+49>:    call   0x401c00 <puts>
   0x00401376 <+54>:    mov    DWORD PTR [esp],0x403024
   0x0040137d <+61>:    call   0x401c00 <puts>
   0x00401382 <+66>:    mov    edx,DWORD PTR [esp+0x1c]
   0x00401386 <+70>:    mov    eax,DWORD PTR [esp+0x1c]
   0x0040138a <+74>:    lea    ecx,[esp+0x1c]
   0x0040138e <+78>:    mov    DWORD PTR [esp+0xc],ecx
   0x00401392 <+82>:    mov    DWORD PTR [esp+0x8],edx
   0x00401396 <+86>:    mov    DWORD PTR [esp+0x4],eax
   0x0040139a <+90>:    mov    DWORD PTR [esp],0x403071
   0x004013a1 <+97>:    call   0x401c08 <printf>
   0x004013a6 <+102>:   mov    edx,DWORD PTR [esp+0x18]
   0x004013aa <+106>:   mov    eax,DWORD PTR [esp+0x18]
   0x004013ae <+110>:   lea    ecx,[esp+0x18]
   0x004013b2 <+114>:   mov    DWORD PTR [esp+0xc],ecx
   0x004013b6 <+118>:   mov    DWORD PTR [esp+0x8],edx
   0x004013ba <+122>:   mov    DWORD PTR [esp+0x4],eax
   0x004013be <+126>:   mov    DWORD PTR [esp],0x40308c
   0x004013c5 <+133>:   call   0x401c08 <printf>
   0x004013ca <+138>:   mov    DWORD PTR [esp],0x403024
   0x004013d1 <+145>:   call   0x401c00 <puts>
   0x004013d6 <+150>:   mov    eax,0x0
   0x004013db <+155>:   leave
   0x004013dc <+156>:   ret
End of assembler dump.
(gdb)

变量ab的地址应该分别在0x22ff4c0x22ff48

(gdb) print &a
 = (int *) 0x22ff4c

(gdb) print &b
 = (int *) 0x22ff48

但是,如果您查看以下 GDB 反汇编输出,地址不是 0x22ff4c & 0x22ff48,而是 0x0040134e & 0x00401356

   0x0040134e <+14>:    mov    DWORD PTR [esp+0x1c],0x3039
   0x00401356 <+22>:    mov    DWORD PTR [esp+0x18],0x12345

我也在 x32dbg 上调试过这个,但也得到了相同的地址。

我在这里很困惑。如果 0x0040134e & 0x00401356 不是变量 a & b 的内存地址,那么它们是什么?

0x0040134e <+14>:    mov    DWORD PTR [esp+0x1c],0x3039
0x00401356 <+22>:    mov    DWORD PTR [esp+0x18],0x12345

不,0x0040134e0x00401356 不是变量 ab 的地址。相反,它们是代码中指令的地址,即 instruction pointer IPab 在这些行中初始化:

0x0040134e <+14>:    mov    DWORD PTR [esp+0x1c],0x3039
0x00401356 <+22>:    mov    DWORD PTR [esp+0x18],0x12345

这里ab的地址分别是esp+0x1cesp+0x18,其中espstack pointer。作为地址,您将获得 0x22ff4c0x22ff48,您可以从中轻松推断出当前 esp0x22ff4c - 0x1c

请注意这个答案忽略了操作系统的底层虚拟内存管理或其他相关的内存管理问题,即在大多数架构上 0x22ff4c0x22ff48 将是虚拟内存地址,而不是真实的RAM 的物理地址。