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本帖最后由 kangels 于 2013-3-19 10:23 编辑 ( I% d' `2 m5 c
- H9 @' ^1 P9 d丢失的类型信息4 Y0 R }8 W0 n) `; `& y. N
猜测
+ m3 m* l% [; n指针变量应该存储两方面的信息：地址，类型信息* ?9 i6 b. E$ m! g8 J. t
9 y2 g1 v7 i1 a; y& N实证6 b+ M: i1 U- D1 Q4 R
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息! z" \, }- g& ^
反汇编一段代码来看看3 D4 S4 Y; a* u
实例1：
3 J% |/ N) O1 a+ Y* I+ P#include <stdio.h> int gi; int * pi; void main(int argc, char **argv) { pi = &gi; *pi = 12; } 1: #include <stdio.h> 2: int gi; 3: int * pi; 4: void main(int argc, char **argv) 5: { 00411370 push ebp 00411371 mov ebp,esp 00411373 sub esp,0C0h 00411379 push ebx 0041137A push esi 0041137B push edi 0041137C lea edi,[ebp+FFFFFF40h] 00411382 mov ecx,30h 00411387 mov eax,0CCCCCCCCh 0041138C rep stos dword ptr es:[edi] 6: pi = &gi; 0041138E mov dword ptr ds:[00417140h],417144h 7: *pi = 12; 00411398 mov eax,dword ptr ds:[00417140h] 0041139D mov dword ptr [eax],0Ch 8: } 004113A3 xor eax,eax 004113A5 pop edi 004113A6 pop esi 004113A7 pop ebx 004113A8 mov esp,ebp 004113AA pop ebp 004113AB ret 0 K. g: H4 {9 ^. A; `
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实例2：
5 Q) o f1 o2 x# P V#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } / s. [6 M; z' b, |( w0 e
1: #include <stdio.h> 2: short gs; 3: short * ps; 4: void main(int argc, char **argv) 5: { 00411370 push ebp 00411371 mov ebp,esp 00411373 sub esp,0C0h 00411379 push ebx 0041137A push esi 0041137B push edi 0041137C lea edi,[ebp+FFFFFF40h] 00411382 mov ecx,30h 00411387 mov eax,0CCCCCCCCh 0041138C rep stos dword ptr es:[edi] 6: ps = &gs; 0041138E mov dword ptr ds:[00417140h],417144h 7: *ps = 12; 00411398 mov eax,0Ch 0041139D mov ecx,dword ptr ds:[00417140h] 004113A3 mov word ptr [ecx],ax 8: } 004113A6 xor eax,eax 004113A8 pop edi 004113A9 pop esi 004113AA pop ebx 004113AB mov esp,ebp 004113AD pop ebp 004113AE ret
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9 n2 O: @! Y' d9 I& ]实例3：
1 B* z. {' G! n#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
+ x9 Z* Q* a( c; ~! o9 T 1: #include <stdio.h> 2: char gc; 3: char * pc; 4: void main(int argc, char **argv) 5: { 00411370 push ebp 00411371 mov ebp,esp 00411373 sub esp,0C0h 00411379 push ebx 0041137A push esi 0041137B push edi 0041137C lea edi,[ebp+FFFFFF40h] 00411382 mov ecx,30h 00411387 mov eax,0CCCCCCCCh 0041138C rep stos dword ptr es:[edi] 6: pc = &gc; 0041138E mov dword ptr ds:[00417140h],417144h 7: *pc = 12; 00411398 mov eax,dword ptr ds:[00417140h] 0041139D mov byte ptr [eax],0Ch 8: } 004113A0 xor eax,eax 004113A2 pop edi 004113A3 pop esi 004113A4 pop ebx 004113A5 mov esp,ebp 004113A7 pop ebp 004113A8 ret
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可以看出指针的类型信息决定了赋值/读取时读/写多少个字节。
% L- E; {4 E- ?+ Z7 G读/写多少字节的信息不是存放在指针变量中，而是放到了与该地址相关的赋值指令中。
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发表于 2013-3-12 20:59:39
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貌似很高深的反汇编
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