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本帖最后由 kangels 于 2013-3-19 10:23 编辑 @$ v, Q3 q0 k* ]) @
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丢失的类型信息0 F) A" \) \# L; L! J( C
猜测
9 M: p" I/ d! ^) Q. x8 `6 z- V% V指针变量应该存储两方面的信息：地址，类型信息
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% K7 s! p( x/ j* D I0 O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 |' ]8 h k3 L% V+ e+ @( L反汇编一段代码来看看2 j# W+ |1 E0 w/ g) K9 q
实例1：
% _: @$ M' e* q# X0 c#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 6 Z% ^+ a: s h
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实例2：0 v* D# w! v1 n# @. r" I i5 a7 b
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
( k3 Z5 u! i% U+ s5 q5 n; m, G; Y2 } 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 y7 A8 F+ ?0 Q5 h* Z! e
: L$ w5 P) K, Q( E实例3：4 X3 y; b1 t2 o! Q
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } ( D, f, ^/ g( Z6 s5 \
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 + @* o. P7 ?/ {. h2 Z @9 s
5 }2 V% p" A- r$ B5 s8 e: H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。
0 x, _; |: I5 c0 {9 R) R6 @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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