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本帖最后由 kangels 于 2013-3-19 10:23 编辑 . e4 s2 V& h8 F7 N$ I6 u
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丢失的类型信息. z; Q- q) Y) N# ]' C8 P9 G4 E
猜测0 a/ [, X3 \4 J' d1 w
指针变量应该存储两方面的信息：地址，类型信息 L+ n2 o5 o6 @% c
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实证# H( Z$ z9 k7 |# ]; |9 o( h
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
4 i# W; n( m: u, Y$ j; J* E7 o反汇编一段代码来看看
# l9 a; Z1 L6 P1 U1 y; k! a实例1：
5 a+ \; k! c' H/ U# {; b#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：+ G2 E2 K; l* N2 K- Q' A
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
; I( y/ {5 w; A( ?1 R8 P 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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实例3：4 ]5 V( Y, a+ E$ Q+ @+ J! i* A4 [
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
/ g- q0 \* j8 I$ K) y 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 1 T8 g; M. N. ^. T5 u7 u
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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。
2 h# r4 c- ?: ], f- S! U+ 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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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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