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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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* C. B6 e+ @& w1 ~( i丢失的类型信息
- J# t8 j! s0 `9 S9 p猜测
: U3 y) r' V) x8 r- x1 _ j指针变量应该存储两方面的信息：地址，类型信息
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1 i7 K! G& n8 V$ @实证* e1 A+ e. [ J% O- @ A
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
1 K+ u' u: a8 H4 `: `% ~反汇编一段代码来看看& U$ S* r; \% T+ V: {* G4 ~
实例1：( q" _$ Z" p: R0 H4 Z2 G$ H, t
#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 8 S3 q2 j' u, k- L) _0 _
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实例2：
5 G8 ^ n& W1 E1 G#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } 4 |; P7 [4 ?' @4 q( E
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：
' a) G- G. o4 ]* t1 l7 Z7 u8 y#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } 3 ^( F1 {7 ]2 M# U3 J- x
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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) h( v2 j/ l4 l; F4 b, @1 B2 B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。
! ^! j8 t; O! Z+ T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。1 _! t9 ]0 n5 ]! \, o3 w- T6 S
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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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