麻烦帮小弟我解释一下这个汇编程序

麻烦帮我解释一下这个汇编程序
#include<iostream>
using namespace std;
int main()
{
012113A0 push ebp  
012113A1 mov ebp,esp  
012113A3 sub esp,0E4h  
012113A9 push ebx  
012113AA push esi  
012113AB push edi  
012113AC lea edi,[ebp-0E4h]  
012113B2 mov ecx,39h  
012113B7 mov eax,0CCCCCCCCh  
012113BC rep stos dword ptr es:[edi]  
int *p=new int [5];
012113BE push 14h  
012113C0 call operator new[] (12110A0h)  
012113C5 add esp,4  
012113C8 mov dword ptr [ebp-0E0h],eax  
012113CE mov eax,dword ptr [ebp-0E0h]  
012113D4 mov dword ptr [p],eax  
delete [] p;
012113D7 mov eax,dword ptr [p]  
012113DA mov dword ptr [ebp-0D4h],eax  
012113E0 mov ecx,dword ptr [ebp-0D4h]  
012113E6 push ecx  
012113E7 call operator delete[] (121101Eh)  
012113EC add esp,4  
return 0;
012113EF xor eax,eax  


}

[解决办法]

Assembly code

#include<iostream>using namespace std;int main(){    012113A0 push ebp      012113A1 mov ebp,esp      012113A3 sub esp,0E4h      012113A9 push ebx      012113AA push esi      012113AB push edi      012113AC lea edi,[ebp-0E4h]      012113B2 mov ecx,39h      012113B7 mov eax,0CCCCCCCCh      012113BC rep stos dword ptr es:[edi]  // 以上是进入函数的压栈操作以及准备工作    int *p=new int [5];    012113BE push 14h            // 将十进制的20入栈    012113C0 call operator new[] (12110A0h)  // (利用上面的20为参数)调用new[] 例程    012113C5 add esp,4            // esp += 4(之前20被压栈，占用4字节，现在释放它)    012113C8 mov dword ptr [ebp-0E0h],eax      012113CE mov eax,dword ptr [ebp-0E0h]  //上面两句很奇怪，就是eax <--- eax (eax的值传给eax)    012113D4 mov dword ptr [p],eax        // 将eax的值，也就是new int[5]返回的指针值传给p    delete [] p;    012113D7 mov eax,dword ptr [p]      012113DA mov dword ptr [ebp-0D4h],eax      012113E0 mov ecx,dword ptr [ebp-0D4h]  // 上面3句，将p传到ecx中    012113E6 push ecx                        // 参数压栈    012113E7 call operator delete[] (121101Eh)  // 调用delete [] 例程    012113EC add esp,4            // esp += 4(之前ecx被压栈，占用4字节，现在释放它)    return 0;    012113EF xor eax,eax        // 设置eax为0        }