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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! w2 s/ \+ w* Z
猜测
/ ?1 R, \* ?# F& c指针变量应该存储两方面的信息：地址，类型信息
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5 b" |9 [$ W9 E* r4 L9 \实证
7 _" D# z j+ v9 U7 P9 x用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息: A! P3 Z* s/ I ^! P
反汇编一段代码来看看3 y1 b7 _. I* K" ?7 U
实例1：) Z9 L; X# B+ c/ T8 q$ A3 m, |
#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：
! K: {1 E1 a- {9 U: K#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } ! K! j; f2 C( T) |& r6 g. r" D
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 ! t9 T3 h5 m+ A+ d) ~0 v! u
1 O9 z) k* ~! e5 w实例3：5 G0 _' ]- {+ W8 a
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
, k1 s/ E, m! {5 d, d# c+ N 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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5 k, |' T7 U$ Z# O/ Q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。
_8 n; e3 h/ F" l8 w- b E$ H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。) q# E, I/ U3 W2 w! o, u1 c; |1 U
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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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