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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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) u$ H1 K, Q) e/ `" G9 N6 [6 r8 `1 _丢失的类型信息
4 m. d+ b% A' c猜测
# o2 d! e! h! C' K7 g. Z% H, P {指针变量应该存储两方面的信息：地址，类型信息" k( y7 s9 o0 k
/ Z7 A0 M+ S$ ]/ ~实证
9 r9 n& `4 Y1 U1 {6 w# E用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息" k# R4 b1 }. H" }$ l
反汇编一段代码来看看" j* p. c5 |" L. c6 Y
实例1：
6 ?, n1 q2 X4 P$ Q8 b#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 % I2 ?7 p% M; f* z6 ~6 w' J; g/ ^( Z
3 E: n: S3 ?6 m5 D7 Y* Q. l: Q& N实例2：
0 [% t8 r% l5 d% F& Y/ O#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
: x0 K- m4 w* @$ i' 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 ) ?4 t/ Q5 {3 y! w4 T8 v
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实例3：
. H: G. w- P% D" g#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
& G5 ]7 O9 W9 U& l' Z* r 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。' c* Z- L: U/ Q& b% `. h8 Q0 X
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。$ y$ _- s9 M+ K) 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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