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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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丢失的类型信息$ O P' ]- _2 _/ t5 N% B* J
猜测5 H( h9 z3 E; o! [7 W" {) a! F
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2 q) n. ~9 S% r8 |+ R0 }
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息2 x7 v* h7 H5 v# ~; m% [1 e" I/ c
反汇编一段代码来看看# q! w( j+ A4 @+ \3 f8 D. X; W
实例1：
+ n1 s' x* `3 l* H0 W+ `3 f* 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
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9 \2 I z4 L) i$ J5 [; a, g实例2：
( H/ U7 E# Y X. t2 q#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } 4 @! G7 i0 b( V2 J: ^
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 , k' h% V2 L2 v/ R6 e
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实例3：( p! J: z% O: N! z
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } , C) G! T" S/ T* Y
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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! X2 y& M# j" J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。
5 M4 l& Z& k& _% |5 s9 h/ j( X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。7 M* A$ T- M) o( u) d) E
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狗仔卡
发表于 2013-3-12 20:59:39
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貌似很高深的反汇编
楼主
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