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4: int main()
5: {
00401010 push ebp
00401011 mov ebp,esp
00401013 sub esp,44h //分配栈内存
00401016 push ebx
00401017 push esi
00401018 push edi
00401019 lea edi,[ebp-44h]
0040101C mov ecx,11h
00401021 mov eax,0CCCCCCCCh
00401026 rep stos dword ptr [edi]
6: int a;//(1)
7: a = 12;//(2)
00401028 mov dword ptr [ebp-4],0Ch
8:
9: return 0;
0040102F xor eax,eax
10: }
4: int main()
5: {
00401010 push ebp
00401011 mov ebp,esp
00401013 sub esp,44h
00401016 push ebx
00401017 push esi
00401018 push edi
00401019 lea edi,[ebp-44h]
0040101C mov ecx,11h
00401021 mov eax,0CCCCCCCCh
00401026 rep stos dword ptr [edi]
6: int a;//(1)
7: a = 12;//(2)
00401028 mov dword ptr [ebp-4],0Ch
8:
9: return 0;
0040102F xor eax,eax
10: }
4: void fun_1(void);//(1)
5:
6: int main()
7: {
00401010 push ebp
00401011 mov ebp,esp
00401013 sub esp,40h
00401016 push ebx
00401017 push esi
00401018 push edi
00401019 lea edi,[ebp-40h]
0040101C mov ecx,10h
00401021 mov eax,0CCCCCCCCh
00401026 rep stos dword ptr [edi]
8: fun_1();//(2)
00401028 call @ILT+5(fun_1) (0040100a)
9: return 0;
0040102D xor eax,eax
10: }
12: void fun_1(void)//(3)
13: {
0040D3F0 push ebp
0040D3F1 mov ebp,esp
0040D3F3 sub esp,44h
0040D3F6 push ebx
0040D3F7 push esi
0040D3F8 push edi
0040D3F9 lea edi,[ebp-44h]
0040D3FC mov ecx,11h
0040D401 mov eax,0CCCCCCCCh
0040D406 rep stos dword ptr [edi]
14: int a;
15: }
int fun (int i);
int main ()
{
int i=1;
fun(i);
}
int fun (int i)
{
return i;
}
00401001 int 3
00401002 int 3
00401003 int 3
00401004 int 3
@ILT+0(_fun):
00401005 jmp fun (00401070)
@ILT+5(_main):
0040100A jmp main (00401020)
0040100F int 3
00401010 int 3
00401011 int 3
00401012 int 3
00401013 int 3
00401014 int 3
00401015 int 3
00401016 int 3
00401017 int 3
00401018 int 3
00401019 int 3
0040101A int 3
0040101B int 3
0040101C int 3
0040101D int 3
0040101E int 3
0040101F int 3
--- e:\study\vc6\cpractice.c ---------------------------------------------------------------------
1: int fun (int i);
2:
3: int main ()
4: {
00401020 push ebp
00401021 mov ebp,esp
00401023 sub esp,44h
00401026 push ebx
00401027 push esi
00401028 push edi
00401029 lea edi,[ebp-44h]
0040102C mov ecx,11h
00401031 mov eax,0CCCCCCCCh
00401036 rep stos dword ptr [edi]
5: int i=1;
00401038 mov dword ptr [ebp-4],1
6: fun(i);
0040103F mov eax,dword ptr [ebp-4]
00401042 push eax
00401043 call @ILT+0(_fun) (00401005)
00401048 add esp,4
7: }
0040104B pop edi
0040104C pop esi
0040104D pop ebx
0040104E add esp,44h
00401051 cmp ebp,esp
00401053 call __chkesp (004010a0)
00401058 mov esp,ebp
0040105A pop ebp
0040105B ret
--- No source file -------------------------------------------------------------------------------
0040105C int 3
0040105D int 3
0040105E int 3
0040105F int 3
00401060 int 3
00401061 int 3
00401062 int 3
00401063 int 3
00401064 int 3
00401065 int 3
00401066 int 3
00401067 int 3
00401068 int 3
00401069 int 3
0040106A int 3
0040106B int 3
0040106C int 3
0040106D int 3
0040106E int 3
0040106F int 3
--- e:\study\vc6\cpractice.c ---------------------------------------------------------------------
8:
9: int fun (int i)
10: {
00401070 push ebp
00401071 mov ebp,esp
00401073 sub esp,40h
00401076 push ebx
00401077 push esi
00401078 push edi
00401079 lea edi,[ebp-40h]
0040107C mov ecx,10h
00401081 mov eax,0CCCCCCCCh
00401086 rep stos dword ptr [edi]
11: return i;
00401088 mov eax,dword ptr [ebp+8]
12: }
0040108B pop edi
0040108C pop esi
0040108D pop ebx
0040108E mov esp,ebp
00401090 pop ebp
00401091 ret