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
}
------解决方案--------------------
#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
}
------解决方案--------------------
- 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 }