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本帖最后由 kangels 于 2013-3-19 10:23 编辑 / ]8 P' r% P5 _# K( r; N8 v
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丢失的类型信息8 @6 E2 X" f! M5 Z. u
猜测' F6 z, }9 k4 Z/ p. s% W
指针变量应该存储两方面的信息：地址，类型信息
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" W5 L# E, H% X D) M7 m实证; V/ p% D3 b5 Y2 g3 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 t9 p. b: ~: P0 p$ z4 k
反汇编一段代码来看看* D- i/ a# P6 b
实例1：! P% H0 M" E& N- O! x
#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 2 m! l6 i9 D* G- d& p
3 v5 X4 a" t c. B实例2：
/ ]. v' D! |& `1 ~#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
* d5 z, I' Z! ?8 {0 n5 T6 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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' x4 k) X* h7 T' g2 H" p* b0 O实例3：: u# M$ R! y R& _9 y+ r
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } 7 F5 F+ Z) ?( J/ J& H
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 ! D4 f+ J( d" U: X: i1 A6 C
* |+ k0 [7 z5 T0 ?) {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。8 `2 }6 J" J) i5 \# L* Q+ {, B
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。' G$ \" M- d6 L ~ z( x* e9 o& H, U
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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
楼主
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