Question

You could have a try. The program below compiles and runs smoothly. The address of variable ex in the constructor is different from that of e, a temporary variable in the catch block. Yet you may notice that the value of ex in line B is passed to e by reference. Could anyone explain what’s going on?

#include<cstring>
#include<iostream>


using std::string;
using std::endl;
using std::cout;

class ThrowException;
ThrowException* TE_ptr;

class ThrowException{

    private:
        string msg;
        int b;
    public:
        ThrowException(string m="Unknown exception",int factor=0) throw(string,const char*);
        ~ThrowException(){        cout<<"destructor get called."<<endl;}
        friend std::ostream& operator<<(std::ostream& os,const ThrowException&TE);
};

ThrowException::ThrowException(string m, int f) throw(string,const char*):msg(m),b(f){
    cout<<"msg="<<msg<<'\n'<<"b="<<b<<endl;
    TE_ptr=this;
    if(b==1){
        string ex("b=1 not allowed.");
        cout<<"The address of e in constructor is "<<&ex<<endl;      //A
        throw ex;
    }
}

std::ostream&operator<<(std::ostream&os, const ThrowException &TE){
    os<<TE.msg<<'\n'<<TE.b<<endl;
}
int main(){

    try{
        ThrowException a("There's nothing wrong.", 1);
    }catch(string &e){             //B
        cout<<"The address of e in first catch block is "<<&e<<endl;        //C
        cout<<"The content resided in the momery block pointed to by TE_ptr is "<<*TE_ptr<<endl;
    }

}

Another problem I would like to ask is when will the destructor of ThrowException object a be called?

Answer 1

The throw expression copies the thrown object to a safe place before leaving the local scope. The language doesn’t say exactly where it is stored, just that this must work somehow (details left to the implementation).