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本帖最后由 kangels 于 2013-3-19 10:23 编辑 3 l/ i/ ^3 M( J
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丢失的类型信息1 K$ A* W7 J4 \0 d
猜测
! b/ {; C n9 b: ~7 f; a8 }指针变量应该存储两方面的信息：地址，类型信息% a. ^+ T5 v% f5 j
2 R L# q5 O" N/ T实证
& S( y. [* o' O& t$ ~1 _用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
8 i- N3 O3 y7 e反汇编一段代码来看看
5 c$ B. M; s# a; P5 ?+ t实例1：+ L+ A3 Y& @% Z1 f0 y
#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：* \/ z3 u; ^: \/ u: f4 H! Y! p3 z
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
/ ^. f* \: r5 I& r) m; E' r- R 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3：5 k- C+ m4 }& M" I) a
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
( p# B$ O/ p6 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 ; j! [6 B4 a: X9 h4 y2 q9 @
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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。1 [; p7 w; L* A+ v% e' E
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。- E7 i* G( t+ |; O' v& K
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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
楼主
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