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
(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
a和b的地址应该分别位于0x22ff4c和0x22ff48(gdb) print &a
$1 = (int *) 0x22ff4c
(gdb) print &b
$2 = (int *) 0x22ff48
但是，如果您查看以下GDB反汇编输出，则该地址不是
0x22ff4c和0x22ff48，而是0x0040134e和0x00401356 0x0040134e <+14>: mov DWORD PTR [esp+0x1c],0x3039
0x00401356 <+22>: mov DWORD PTR [esp+0x18],0x12345
我也已经在x32dbg上调试了它，但是也得到了相同的地址。
我在这里很困惑。如果
0x0040134e和0x00401356不是变量a和b的内存地址，那是什么？ 最佳答案
0x0040134e <+14>: mov DWORD PTR [esp+0x1c],0x3039
0x00401356 <+22>: mov DWORD PTR [esp+0x18],0x12345
不，
0x0040134e和0x00401356不是变量a和b的地址。相反，它们是代码中指令的地址，即instruction pointer IP。在以下几行中初始化了a和b:0x0040134e <+14>: mov DWORD PTR [esp+0x1c],0x3039
0x00401356 <+22>: mov DWORD PTR [esp+0x18],0x12345
此处
a和b的地址分别是esp+0x1c和esp+0x18，其中esp是stack pointer。作为地址，您将获得0x22ff4c和0x22ff48，从中可以轻松推断出当前esp是0x22ff4c - 0x1c。请注意，此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，在大多数体系结构中，
0x22ff4c和0x22ff48将是虚拟内存地址，而不是RAM的实际物理地址。关于c - 汇编:C编程中变量的内存地址，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/44867145/