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本帖最后由 kangels 于 2013-3-19 10:23 编辑 7 s" i! \+ G1 }& i/ k$ Q
9 _3 {4 ~& \( H丢失的类型信息+ H+ f2 D5 |/ F7 N! M$ K
猜测 S; ^- E7 k) D2 c) n0 c
指针变量应该存储两方面的信息：地址，类型信息: a2 ^1 I( } ~4 f
% v+ t" S+ A; k5 D' n. |; d. F, P实证5 G! H4 {, m' {) J; C
用sizeof()，float *,int*,double *....都是4字节没有存储类型信息只有地址信息' x D9 B5 Y0 n' A. d6 Y
反汇编一段代码来看看3 b. C4 \$ z' M$ {5 [0 `$ {
实例1：+ S, v- k3 c& y& \8 I
#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 , U+ K U: W8 ?! r% R0 e+ y
v& b. u: u; [$ n1 G+ |# P; Q; _. K实例2：
, b( _! c+ i$ N8 S1 M#include <stdio.h> short gs; short * ps; void main(int argc, char **argv) { ps = &gs; *ps = 12; } - L( W5 y& z7 B6 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 5 F0 X- d1 o4 O+ t
, Z6 h3 E) X W6 w" c& i实例3：) W' w: b9 I" P# m @+ P0 N" u
#include <stdio.h> char gc; char * pc; void main(int argc, char **argv) { pc = &gc; *pc = 12; }
$ h6 P! R7 Y3 U& x1 b: m 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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1 N, Z; f2 [9 j9 z: S) T4 j/ q: a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。2 G8 v1 x9 t& K; \8 s5 F1 D- Y1 m
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。# j' k& s4 I3 L2 G3 t' q
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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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