当C++函数的返回类型为一个类时

仅仅是测试一下，当函数返回类型为一个类时的情况。

#include<iostream>using namespace std;class A{public: int i; int *n; A() // 构造函数 { n= new int; *n = 101; } ~A() // 析构函数 { cout<<"Destructor"<<'/t'<<*n<<endl; cout<<this<<endl; delete n; } A& operator =(const A& a) { cout<<"Copy"<<endl; if(this == &a) return *this; this->i = a.i; *(this->n) = *(a.n); return *this; } void print() { cout<<i<<endl; cout<<*n<<endl; } };A GetA(){ A a; a.i =3000; *(a.n)=3001; return a;}int main(){ A *a = new A(); *a = GetA(); a->print(); delete a;}

<script type="text/javascript"><!--google_ad_client = "pub-3527323059587280";/* 468x60, 创建于 08-10-10 */google_ad_slot = "6038158497";google_ad_width = 468;google_ad_height = 60;// --></script>