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本帖最后由 kangels 于 2013-3-19 10:23 编辑 - q* _# T' C6 I1 N5 p
2 \8 E/ n; J# c* P丢失的类型信息
# l5 z9 }- W7 ]% [! ]( P3 _猜测- C6 Y) @+ |/ p& V- [" q
指针变量应该存储两方面的信息：地址，类型信息3 @" V7 U- Y S
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实证5 J* B0 f; u7 c' Y' e
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息* u5 [& j+ H% S
反汇编一段代码来看看
# l7 ~. o" c" i& c3 }实例1：
; y+ S) e* [' K" R+ p& O#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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- G/ H% S' k3 H实例2：
% R" K9 I% w3 [. W4 ^#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
6 o) m4 K: N ^' I1 C 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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实例3：
# @. m0 _7 {- g# R: ~0 W2 i M#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
- E$ }& P. T% Q# x( L1 k 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 9 d0 V0 T5 s3 |4 e: y
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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。
! E. W! R T- ]2 L& [/ l* 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。2 g J% z- u, C0 X
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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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