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本帖最后由 kangels 于 2013-3-19 10:23 编辑 4 F/ ~* i" H7 q9 f7 i. S7 j: q; J
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丢失的类型信息$ c' |9 R( Q; ?1 ^+ S& z5 ?% H
猜测
! h" D z" ?1 `/ m5 F& o8 A指针变量应该存储两方面的信息：地址，类型信息$ ?; q6 L1 W( f! W# \. K8 ^
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实证
4 \. n& Y2 ~6 S% w用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息% n6 B) u& j$ f% N: G
反汇编一段代码来看看- |/ b) ~ z0 x9 \5 q
实例1：
e. u1 a. M' E9 G#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2：8 s9 K" } r/ F9 G q
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
0 R0 A. F1 B2 c5 e6 T/ O 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 : g4 s S+ ?; D$ W2 G: v) \
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实例3：
+ D% a( _4 a+ `9 s) D$ v#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } . F# ?2 |! `7 s b$ d# t% q' 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 7 v% g/ p' u7 h: N* x
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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。
9 K. {6 A! k5 d, K2 c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。7 f6 ~3 z+ ^+ f9 {% m
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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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