麻烦帮我解释一下这个汇编程序

[LearnITing]

#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


}

[程序员小迷]

#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
	
	
}