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本帖最后由 kangels 于 2013-3-19 10:23 编辑 5 V% L, W! c. A0 ^: `1 c
8 z# k2 r: L7 L. l1 A- o丢失的类型信息: z: ]; {0 v# B- f# G
猜测
9 s" |5 f2 g# ~1 q- m6 z指针变量应该存储两方面的信息：地址，类型信息+ I, Q1 T/ F+ h& Q+ r @% p; M; g
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实证
E% [6 i/ h3 R7 Z2 N' N$ D- S6 t用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
% J6 Z4 A7 I: @5 }+ Q反汇编一段代码来看看
4 ]0 K! n, C$ ]) Z( F4 U" G实例1：1 l) q7 A$ P! R0 b' 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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实例2：
4 d: \) r, e7 J$ ^0 Q+ v#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
1 P( c6 S+ Z9 H t- T% K8 ~/ V3 a 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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2 N6 I& l& g& p4 I0 `实例3：
/ h( q" H: m* P#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } ' F; J6 h+ x* I1 D8 G
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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0 D/ {3 p# m5 K% p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。
: Z; E" ]: B; M0 _" }2 ]# e, I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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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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