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本帖最后由 kangels 于 2013-3-19 10:23 编辑 3 z( y7 q1 O6 |& x G" C, t; }9 Y
+ R! j) }% k' L0 F/ T1 }7 ]丢失的类型信息, l0 s+ t+ |, [+ f) K J7 D
猜测
8 P2 L6 @+ q4 c指针变量应该存储两方面的信息：地址，类型信息
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用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
1 u3 ~! U! g5 |, S" L* p$ ~" w7 ]反汇编一段代码来看看1 U7 r5 U7 |0 |
实例1：
* v3 q3 N2 o3 d2 ^7 ~; D8 k2 A1 y0 q#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 3 H$ D0 l' |; h( @3 m; E
6 p4 v7 u8 ?7 ^实例2：
$ L; D& n: M9 A#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
+ h8 X: W* k5 d! I+ }3 a 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 : M, @2 m2 @0 o+ V5 w
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实例3：# k: u& n: ^& f
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
! m' e: M5 l( g- V/ L# Z 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 + l, m! U- q1 g. j" K
, O/ [; j7 _( O: K4 z% W' C+ _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。: s. q; Y5 C. N+ L
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。" m" x" ~) Y, B3 o+ n6 g0 `' K7 P
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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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