c++ - 为什么抛出异常时析构函数调用两次？

Question

我很困惑为什么在抛出异常时会调用两次析构函数，调用的是哪一点？？

#include <iostream> 
using namespace std;
class base
{
 public:
     base(){cout<<"constructor called"<<endl;}
     ~base(){cout<<"destructor called"<<endl;}
};
void fun()
{
     throw base(); //<=- Create temp obj of base, and throw exception

}
int main()
{
    try
    {
        fun();
    }
    catch(...)
    {
        cout<<"handle all exception"<<endl;
    }

}

输出如下

constructor called
destrctor called
handle all exception
destuctor is called

但是当我添加复制构造函数时，它从未调用过，但析构函数只调用了一次所以发生了什么？？？？

#include <iostream> 
using namespace std;
class base
{
 public:
     base(){cout<<"constructor called"<<endl;}
     ~base(){cout<<"destructor called"<<endl;}
     base (base &obj){cout<<"copy constructor called"<<endl;}
};
void fun()
{
     throw base(); //<=- Create temp obj of base, and throw exception
}
int main()
{
    try
    {
        fun();
    }
    catch(...)
    {
        cout<<"handle all exception"<<endl;
    }

}

输出:

constructor called
handle all exception
destrctor called

Answer 1

编译器可以根据需要多次复制您的异常对象。析构函数被调用两次，因为有一个拷贝。