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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, G k6 o8 x8 z8 Y l
猜测
+ c9 A3 `" O5 h指针变量应该存储两方面的信息：地址，类型信息
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实证' d0 ~. \, d' b. Y% d
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息( C% t( G9 R" B' P4 E8 x+ N) k
反汇编一段代码来看看
% V# F5 R. V- `+ W3 E实例1：. T/ w( H7 N8 U9 r( @/ }5 X
#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 ; K/ k5 s# Y/ \
' r/ q E6 f( t+ k0 N实例2：
' k; t4 a- M# S) f1 ]#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } , i3 l+ G+ ~4 @% 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 # Y6 H- W! s6 ~) d1 r7 C* a
, ~% m/ H, d! t% T. z" o7 n实例3：) z/ n# e9 J8 a! M. ?! n
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } " ^0 C1 {, g y& Q$ [1 x5 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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/ X% n5 D U3 |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。: A$ n( D; C; j' ]$ S" i. C5 }$ s
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