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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$ n. `2 N* s9 p2 ]. y9 S9 e( b# F
猜测. }6 ~. L3 y& w' _! o' }# O/ I
指针变量应该存储两方面的信息：地址，类型信息0 }- c: N5 U% ~9 C1 ~2 r7 O' ?: d
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实证; J. o- c8 o; m7 j' s( V
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息, ^/ S7 S# I1 l* P" b; Z$ C
反汇编一段代码来看看
4 i4 C5 Y) g4 w/ F( u实例1：3 b9 x9 H+ @. b0 B) P W
#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：
& W% O9 F# z+ h; u+ J; \+ R) {#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } W+ c- J! ^, A# A8 n
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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4 g% W2 a9 d% x1 Q- j! m实例3：
9 B4 ~* q F2 M& i4 ^' m#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } 9 k8 I7 h& w* i- f
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 A7 W; g( x" x' C- Z; L/ f
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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。
' b! i6 g: A* Y3 j( O4 l$ m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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狗仔卡
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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