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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; e" L1 x- ?; _" V( d; n
猜测1 `' ?$ R- l' r; S, e& x) M( m
指针变量应该存储两方面的信息:地址,类型信息
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( l5 p9 W8 B1 W8 I实证
1 o- x0 p% T# e0 C \1 D9 r用sizeof(),float *,int*,double *....都是4字节没有存储类型信息只有地址信息
5 b0 z- a& g3 T1 l反汇编一段代码来看看9 ?! H: e1 m' I) l- _
实例1:, p1 R2 q' }6 ^' d. E7 T* _% j
#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# [9 k# e+ r" M4 J
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实例2:+ O. Q$ J; |: d k5 |
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
8 q7 ~& l- y1 G2 j- l% v 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 ) y& z8 S/ y. h1 z4 E
3 g/ L6 G- A( G, I2 @实例3:1 [2 B! v2 j) L. _( n. j( x; I
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
* m( L5 K9 ?9 W' ~ 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 8 z/ r6 s. d$ Z7 f6 i" ~: [3 Y3 r
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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。
# H/ h3 _ X! c! W7 X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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