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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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; `! y8 m: [/ v& O* n9 d2 _: N" V/ H% |1 Y丢失的类型信息9 s% P) i" k/ [8 ]9 N O i, M
猜测, u7 s2 a; A& [
指针变量应该存储两方面的信息：地址，类型信息- e5 t; h; s7 ^/ k( A, W& l
4 ~+ f! Q$ J2 ^" k" p实证
* g& m7 F# a+ V h+ B用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息- B; O3 p, @8 J/ B' ]- a h
反汇编一段代码来看看
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1 p& i N( E. o#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：
- E" _. P1 b- ]* u) b#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
+ G+ B8 k1 W4 w, d/ t' ^: C+ J 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 % b( h: d; S, ~. ?6 G4 V M
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实例3：
! k/ z8 m( Y0 h0 p' Y0 _. E$ v+ l#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
2 n3 N$ Q! s* ]$ Q 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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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。$ N) Z0 K3 C) M
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。 V( b- C( k# {: z7 m2 {$ X
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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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