Question

I don’t understand why the output of this program is Second method instead of First Method…

#include <iostream>

template <class T>
void assign(T& t1,T& t2){
    std::cout << "First method"<< std::endl;
}

template <class T>
void assign(T& t1,const T& t2) {
    std::cout << "Second method"<< std::endl;
}

class A
{
public:
    A(int a):_a(a){};
private:
    int _a;
    friend A operator+(const A& l, const A& r);
};

A operator+(const A& l, const A& r) {
friend A operator+(const A& l, const A& r);return A(l._a+r._a);
}

int main ()
{
    A a=1;
    const A b=2;
    assign(a,a+b);
}

However, when I change my main function to this:

int main ()
{
    A a=1;
    const A b=2;
    A c=a+b;
    assign(a,c);
}

The output is First method.
Any ideas?

Answer 1

assign( a, a+b );

The result of a + b is an rvalue expression of type A that creates a temporary and you cannot bind it to a non-const reference, so it picks up the const overload as you are allowed to bind a const reference to a temporary.

assign( a, c );

In this case, the subexpression c is an lvalue expression and you are allowed to bind a non-const reference. In this case, because the non-const version is a perfect match with T=A it is preferred over the const overload that would require a conversion in the second argument from an lvalue of type A to const lvalue of type A.