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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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2 W) y0 m( ]/ h0 K' ]. [8 z- e# x6 s9 d丢失的类型信息
1 d- {( i+ E/ i K9 G猜测
7 d0 K" V$ C. Z! p$ ~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
- y2 u5 P- c; C( |3 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
0 s7 M$ F1 \1 O' a反汇编一段代码来看看
. v+ i* k$ L. x' E实例1：- Y6 {+ Y: x0 k% \6 n/ ?
#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 9 l$ B, ?. n1 l* M8 x7 i
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实例2：3 V0 ~; b" {6 J3 \" R& c9 x5 b
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } ! D( V- ]! B& d6 z( c2 H) ^
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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: d7 A: i. g8 _% ]% Q" I& Z8 z实例3：
& m& i! x% }, {8 {& h" r#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
, h* C, s; b3 o- ~. }0 x! Q1 \ 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 , T& A% k) ], @3 {
7 @# a* n5 F: s" M/ @; h6 {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。" }4 Q; e9 N9 e* C* [# f
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。+ o0 }* y# W: s" I0 N
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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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