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本帖最后由 kangels 于 2013-3-19 10:23 编辑 + e+ |4 h2 T9 H
7 M, C% D1 V8 d9 j丢失的类型信息/ d1 B. \! N+ P8 _# B
猜测
$ t l6 k4 Q- i( s( S8 J指针变量应该存储两方面的信息：地址，类型信息4 [5 W6 k0 e& K3 k; k
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实证
8 W3 @" \; v, T' h0 [8 V6 r4 V7 m' ?# 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: p0 K5 R" @- V3 U0 y
反汇编一段代码来看看
. n* T' x& _3 p% U; H' g) H实例1：& I9 l b" m; S1 e2 |+ l* Z
#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2：4 J7 Z5 P8 M& f' }) ? J
#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } : s" g" w5 V) y9 C+ Q! o
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 % ~0 s3 \# [! f. h3 ]
6 l# S7 O% R8 B) ], b实例3：- ~+ h- `* g3 S% E2 L
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
# Q1 Q7 B& I2 T) [& L 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 . ~+ }, z$ P% ^; v1 e) |
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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。
w- z7 A$ A! c A9 U& K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。( o+ \) c2 b+ Z, p6 J& y! t1 Q& c
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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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