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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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M9 ?: q; N! Y) @1 \2 S丢失的类型信息+ y0 L* ?+ s3 k7 U; s9 U3 A
猜测6 p! ^1 ?0 T8 X1 y: j- c' 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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实证& m/ x$ Y- { U4 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
$ j; j! h2 R! ~7 b- n8 g反汇编一段代码来看看- s) u; F2 E! n% P! Q0 R
实例1：
2 o4 h) d3 @, U. c" b$ `& Q6 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 0 l0 I- N5 s( j& Z: E, s
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实例2：
( ~; C; M5 h5 z% H2 v* U#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
0 {# n6 R/ V& j, r: w 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 a. f& m& V6 V1 A1 v
^, C2 w; k! @$ f5 L8 X/ D8 h1 l3 P实例3：
7 T6 l# F# E% v% p9 @! M, X e% W#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
; s% J3 W W# @' M0 M3 } 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 + Y3 D5 d# I& i5 i9 `
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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。% q+ O9 D/ y) U/ V
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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