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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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/ W/ z- N' k$ n, s3 _5 F# Y, h丢失的类型信息, `* @0 K# ^/ K3 z. ^" r
猜测
8 J! h* m) o5 }, i1 ?# L( {指针变量应该存储两方面的信息：地址，类型信息
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实证
) k9 p6 @ t5 ]) n* g' X9 D用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
4 [( h2 Q& w n% F3 e$ ?1 u" H反汇编一段代码来看看
, e0 n7 i2 @4 v实例1：
W/ B% x1 S: X8 j) e- a* V#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 - F* B1 e( o) Q) b
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实例2：% b! j; R' z& U4 V
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } $ t7 i2 y3 a8 T ?# E2 w1 x
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 8 R: q1 G: }2 P8 C' b1 @1 }
, c& t/ W+ c7 t) J实例3：
4 `' @5 }' Y1 ?$ a/ [* _#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } % K1 h4 h7 q5 D% J6 C: M* ]
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 / k+ D9 |2 k x0 V1 F
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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。
; P; t7 |4 G* ]+ Q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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