Question

In the following codes, I illustrate an example of operator overloading:

#include <iostream>
using namespace std;

template <typename T>
class A
{
public:
    A()   {};
    A( T &obj) {value = obj;};
    ~A() {};
    T value;
    template <typename E>
    A<T>& operator = (const A<E> &obj)
    {
        cout<<"equal operator"<<endl;
        if(this == &obj)
            return *this;

        value = obj.value;
        return *this;
    }

};



int main()
{
    int temp;
    temp = 3;
    A<int> myobjects(temp);
    cout<<myobjects.value<<endl;

    temp = 7;
    A<int> yourobjects(temp);
    yourobjects = myobjects;
    cout<<yourobjects.value<<endl;



    return 0;
}

However, when I debug this program, I find that the main program does not invoke the equal operator overloading function. However, if I change the equal operator in the following way:

   A<T>& operator = (const A<T> &obj)
    {
        cout<<"equal operator"<<endl;
        if(this == &obj)
            return *this;

        value = obj.value;
        return *this;
    }

It will work. Do you have any ideas why the initial function does not work?

Answer 1

Your template version of assignment operator does not suppress the generation of compiler-provided non-template copy-assignment operator for your class. The compiler will implictly declare and define a copy-assignment operator with the following signature

A<T>& operator =(const A<T>&);

In the overload resolution process for copy-assignment the compiler-provided version wins (since it is more specialized).

Your template version of assignment operator will only be considered for conversion-assignment. E.g. if at some point you’ll want to assign A<int> object to A<double> object, your template version of assignment operator will be used. But when you assign A<int> to A<int> your operator is ignored since the compiler-declared version is a better match.

When you declare your own version of copy-assignment with

A<T>& operator =(const A<T>&);

signature, it suppresses the compiler-generated one. Your version gets used.

This means that if you want to have your own copy-assignment operator as well as template conversion-assignment operator, you need to explicitly implement both in your class.

P.S. As ‘@Cheers and hth. – Alf’ correctly noted, your template version of assignment operator is not even valid in general case. Pointers this and &obj generally have different unrelated types. You are not allowed to compare pointers of different unrelated types.