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本帖最后由 kangels 于 2013-3-19 10:23 编辑
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' j! _! }$ y- x7 C2 Y& Q6 Q. u1 X2 {丢失的类型信息- H/ p) G' H9 y7 u8 r
猜测3 _) V! P) F) B- r
指针变量应该存储两方面的信息：地址，类型信息
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k9 G0 q% c0 D实证) I( W! P( J4 r6 b3 a% ^' _% b+ X
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息
9 l; \2 E' g0 R3 h0 L2 F3 V反汇编一段代码来看看# \& M/ t( N2 S& H! o& p3 D8 b5 J
实例1：' i7 C; L9 c2 T0 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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! j8 e2 O- R/ p0 Q实例2：
5 n4 ~. _6 E1 I/ z0 `7 f#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; }
3 h, s1 i, \# \ s' h) P5 F 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 7 _' j6 q# `7 u g5 V/ I4 ~
# [4 d1 G0 Y2 f# w实例3：
. U# y) x: n1 y2 ~! b#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
& [0 r7 S- X" A T: O; L, J/ X4 @ 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 ; u, J% l* A7 w) C& B. p4 S
$ E' w* ]. s6 p! A# u, }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。
l- K% A: _2 z7 f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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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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