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本帖最后由 kangels 于 2013-3-19 10:23 编辑 . t# |6 u( n) T9 @
. h3 p; |# h$ D& d' a丢失的类型信息
y4 U! c+ v: c, C猜测
6 E1 W+ e2 _4 L1 B: w$ W9 z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
% T _" K% q) m f, h/ k用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
' \% J* Y# `: Q; _8 X反汇编一段代码来看看
( `/ M; K6 _: X/ R4 y3 f实例1：
0 m! o. o* E$ Y( q4 C% p) I" Z#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：
0 L0 J$ [; U. V0 y+ W: V, L#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
7 E/ t1 e' C: |; q0 z0 q 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 - }1 @( d; O6 F+ ?# V- v) d: f
; p" c+ d! m% Z: ]0 W" m5 ]7 S实例3：2 S% @; n0 t8 C+ o. j
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } 1 G7 \, R' J: C* _; s- s
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。
; i4 N- r3 C* @0 i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。; J. D) g5 F! H4 A1 u$ k
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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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