CODEWITHSUNDEEP
c++operator-overloading
Netap
Netap

Reputation: 193

C++ using operator int() instead of operator+

I'm trying to understand why the operator int() is invoked instead of the defined operator+

class D {
    public:
        int x;
        D(){cout<<"default D\n";}
        D(int i){ cout<<"int Ctor D\n";x=i;}
        D operator+(D& ot){ cout<<"OP+\n"; return D(x+ot.x);}
        operator int(){cout<<"operator int\n";return x;}
        ~D(){cout<<"D Dtor "<<x<<"\n";}
};

void main()
{
    cout<<D(1)+D(2)<<"\n";
    system("pause");
}

my output is:

int Ctor D
int Ctor D
operator int
operator int
3
D Dtor 2
D Dtor 1

Upvotes: 10

Views: 4203

Answers (1)

Arunmu
Arunmu

Reputation: 6901

Your expression D(1)+D(2) involves temporary objects. So you have to change you signature of operator+ to take by const-ref

#include <iostream>
using namespace std;

class D {
    public:
        int x;
        D(){cout<<"default D\n";}
        D(int i){ cout<<"int Ctor D\n";x=i;}
        // Take by const - reference
        D operator+(const D& ot){ cout<<"OP+\n"; return D(x+ot.x);}
        operator int(){cout<<"operator int\n";return x;}
        ~D(){cout<<"D Dtor "<<x<<"\n";}
};

int main()
{
    cout<<D(1)+D(2)<<"\n";
}

It prints:

int Ctor D
int Ctor D
OP+
int Ctor D
operator int
3
D Dtor 3
D Dtor 2
D Dtor 1

The operator int is invoked while finding the correct overload for printing it out to cout.

Upvotes: 10

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