C++的class由C的srtuct變化而來，先來看看兩個地方有什么區別:
1.C++代碼
1
#include "iostream"
2using namespace std;
3
4struct A
5
{
6
int a;
7 void display(int s);
8
};
9
10void A::display(int s)
11{
12 a = 1;
13 cout<<"this is in A:"<<s<<" "<<a<<endl;
14}
15
16class B
17{
18public:
19 int b;
20 void display(int s);
21};
22
23void B::display(int s)
24{
25 b = 2;
26 cout<<"this is in B:"<<s<<" "<<b<<endl;
27}
28
29void main()
30{
31 A a;
32 a.display(sizeof(a));
33
34 B b;
35 b.display(sizeof(b));
36}
37
2.匯編代碼：
1.debug編譯
1.text:00401820 main proc near ; CODE XREF: _mainj
2.text:00401820
3.text:00401820 var_48 = dword ptr -48h
4.text:00401820 var_8 = dword ptr -8
5.text:00401820 var_4 = dword ptr -4
6.text:00401820
7.text:00401820 push ebp
8.text:00401821 mov ebp, esp
9.text:00401823 sub esp, 48h
10.text:00401826 push ebx
11.text:00401827 push esi
12.text:00401828 push edi
13.text:00401829 lea edi, [ebp+var_48]
14.text:0040182C mov ecx, 12h
15.text:00401831 mov eax, 0CCCCCCCCh
16.text:00401836 rep stosd
17.text:00401838 push 4
18.text:0040183A lea ecx, [ebp+var_4]
19.text:0040183D call j_A__display
20.text:00401842 push 4
21.text:00401844 lea ecx, [ebp+var_8]
22.text:00401847 call j_B__display
23.text:0040184C pop edi
24.text:0040184D pop esi
25.text:0040184E pop ebx
26.text:0040184F add esp, 48h
27.text:00401852 cmp ebp, esp
28.text:00401854 call __chkesp
29.text:00401859 mov esp, ebp
30.text:0040185B pop ebp
31.text:0040185C retn
32.text:0040185C main endp
33
2.release編譯
.text:00401140 ; int __cdecl main(int argc,const char **argv,const char *envp)
.text:00401140 _main proc near ; CODE XREF: start+AFp
.text:00401140
.text:00401140 var_8 = dword ptr -8
.text:00401140 var_4 = dword ptr -4
.text:00401140 argc = dword ptr 4
.text:00401140 argv = dword ptr 8
.text:00401140 envp = dword ptr 0Ch
.text:00401140
.text:00401140 sub esp, 8
.text:00401143 lea ecx, [esp+8+var_8]
.text:00401147 push 4
.text:00401149 call sub_401000
.text:0040114E push 4
.text:00401150 lea ecx, [esp+0Ch+var_4]
.text:00401154 call sub_4010A0
.text:00401159 add esp, 8
.text:0040115C retn
.text:0040115C _main endp
3.輸出結果
this is in A:4 1
this is in B:4 2
4.結論
1.struct和class沒有任何區別，他們在代碼段有一個"模板"
2.對象占用的4個字節是int的大小
3.函數在代碼中定義，由編譯器決定調用誰
posted on 2008-03-02 21:16
margin 閱讀(482)
評論(0) 編輯 收藏 引用 所屬分類:
C/C++ 、
逆向工程