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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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" O5 R& [, Z+ ~3 |/ P$ m! [0 h( H丢失的类型信息
6 l: ~3 e' V2 @' e+ P* V猜测
. Y I9 H2 c0 P( M2 G& H指针变量应该存储两方面的信息：地址，类型信息9 j, E1 c) `! Z0 d% n" L
9 j7 V' j6 L6 `7 v/ Q) d实证
& G. t9 {% F) H用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息' a5 \3 X) t1 ^6 |
反汇编一段代码来看看( U* e6 K3 y8 a- ^
实例1：
9 G, p5 y1 R, K: J' @0 O2 S#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 & O" ?" f& j" N
5 z* \/ Z( R, B0 _4 U9 [实例2：
* U: ^2 f+ I# h( |; q% _#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } + t7 @2 P- T) v" j6 [5 L# 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 4 `/ E- ]* ~& V: h2 N2 \* I% D
7 e8 u& T0 y. N实例3：
6 {7 n Y9 L$ W0 p$ f#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } 0 Z2 M3 t* @! A% 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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/ S/ p8 ^# V { R8 h n4 b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。
3 j" g$ Z5 p; ?# B* Q4 T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。- ~" h# \% |" D7 y4 Z* N
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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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