析构不匹配的问题

[sybtjp]

关于自定义异常：
#include <iostream>
using namespace std;
class MyException{
public:
MyException(){cout < <"construction" < <endl;}
void show(){cout < <"...MyException..." < <endl;}
~MyException(){cout < <"~construction" < <endl;}
};
void fun(){
throw MyException();
}
int main(int argc, char *argv[]){
try{
fun();
}catch(MyException e){
e.show();
}
return 0;
}
对于此函数为什么会是这样的结果呢？
construction
~construction
~construction
...MyException...
~construction
~construction
不是说构造了几次就析构几次么，这里怎么不一样呢，

[zclever]

//MinGW
#include <iostream> 
using namespace std; 
class MyException{ 
    public: 
        MyException(){cout <<"construction" <<endl;} 
        void show(){cout <<"...MyException..." <<endl;} 
        ~MyException(){cout <<"~construction" <<endl;} 
}; 
void fun(){ 
    throw MyException(); 
} 
int main(int argc, char *argv[]){ 
    try{ 
        fun(); 
    }catch(MyException e){ 
        e.show(); 
    } 
    return 0; 
} 

结果：
/****************************/
construction
...MyException...
~construction
~construction


Terminated with return code 0
Press any key to continue ...
/****************************/

[zclever]

//MinGW
#include <iostream> 
using namespace std; 
class MyException{ 
    public: 
        MyException(){cout <<"construction" <<endl;} 
        void show(){cout <<"...MyException..." <<endl;} 
        ~MyException(){cout <<"~construction" <<endl;} 
}; 
void fun(){ 
    throw MyException(); 
} 
int main(int argc, char *argv[]){ 
    try{ 
        fun(); 
    }catch(MyException e){   //这里调用的不是你提供的默认构造函数，而是它的默认拷贝构造函数
        e.show(); 
    } 
    return 0; 
} 



//结果：
/****************************/
/*construction
...MyException...
~construction
~construction


Terminated with return code 0
Press any key to continue ...*/
/****************************/

[skyering]

construction
~construction
...MyException...
~construction
~construction

我的运行结果。奇怪，怎么会不一样呢 ，我的是vs2008

[太乙]

#include <iostream> 
using namespace std; 
class MyException{ 
public: 
	MyException(){cout <<"construction" <<endl;} 
	MyException(const MyException& me)
	{
		cout<<"fff"<<endl;
	}
	void show(){cout <<"...MyException..." <<endl;} 
	~MyException(){cout <<"~construction" <<endl;} 
}; 
void fun(){ 
    throw MyException(); 
} 
int main(int argc, char *argv[]){ 
    try{ 
        fun(); 
    }catch(MyException e){ 
        e.show(); 
    } 
    return 0; 
} 

[青葱岁月不可追]

……

[lyghe]

try{ 
        fun(); 
    }catch(MyException e){ 
        e.show(); 
    } 
    return 0;

这里进入catch的时候重新建立了一个MyException对象，但是是通过MyException的拷贝构造函数构造的，而不是默认构造函数。
用这个，就不同啦：

try{ 
        fun(); 
    }catch(MyException & e){ 
        e.show(); 
    } 
    return 0;

一般来说，用引用就可以了。

[ilovezc]

#include <iostream> 
using namespace std; 
class MyException{ 
    public: 
        MyException()
		{cout <<"construction" <<endl;} 
		MyException(MyException &)
		{cout <<"copy construction" <<endl;} 
        void show()
		{cout <<"...MyException..." <<endl;} 
        ~MyException()
		{cout <<"~construction" <<endl;} 
}; 
void fun(){ 
    throw MyException(); 
} 
int main(int argc, char *argv[]){ 
    try{ 
        fun(); 
    }
	catch(MyException &e){ 
        e.show(); 
    } 
    return 0; 
}

得到的结果是

construction
...MyException...
~construction

为什么没有copy construction输出呢

[lyghe]

因为用了“引用”，这个和函数的传入参数是一样的。