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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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3 x) A; Q/ o( H* j. b" y2 T丢失的类型信息
0 C# G X" Z; a4 C/ c$ U猜测
+ s) e! B& T; B$ ~指针变量应该存储两方面的信息：地址，类型信息
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3 J5 K$ M y4 ~6 i1 t B0 s* q实证
1 Y1 ^0 [: `$ {9 t. N5 @用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息; Z6 M. B, R' f
反汇编一段代码来看看& {: M/ z- m; [0 P: R* @
实例1：7 L- _/ G8 ]8 M5 R4 j
#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：
0 v# `2 s8 w! S" i2 T: L#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } / X7 ]$ h5 o) a+ |
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 ! G4 h. G8 J! L9 ~: J* d+ B! G
. L) o, t5 n a2 B$ ]( j' g. N实例3：
4 u8 j: D# y, u' W4 j7 @#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; } 6 F2 A3 `1 k. N4 z, U$ L/ r1 O
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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) U }/ X& w, r0 a6 Y; [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。 N; {& x. N, g9 w; H0 S
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。/ A8 l5 \/ Q4 E9 J+ t2 ~! 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
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