Templated copy assignment operator

2
votes

As per cppreference, a copy assignment operator should non templated :

A copy assignment operator of class T is a non-template non-static member function with the name operator= that takes exactly one parameter of type T, T&, const T&, volatile T&, or const volatile T&"

But in this sample program, i wrote a templated assignment operator, there is no compilation issue and it infact is called(and not default implicitly generated).

template<typename T1>
class Sample
{
public:
        T1 a;
        Sample(T1 b)
        {
                a=b;
        }

        template<typename T2>
        void operator  = (T2& obj2)
        {
                cout<<"This wont be called";
                (*this).a=obj2.a;
        }

};


        Sample<int> obj1(2);
        Sample<int> obj2(3);

        obj2=obj1;

Output:

This wont be called

Am i misunderstanding something?

2
c++templates

2 Answers

2
votes
template<typename T2>
    void operator=(T2& obj2) { ... }

defines a operator= function but it is not the copy-assignment operator as per the standard. To qualify as a copy-assignment operator, it has to be:

void operator=(Samble& obj2) { ... }

or better 

Sample& operator=(Sample const& obj2) { ... }

Given your code, you may use:

Sample<int> a;
Samble<double> b;
b = a;

If you had the real copy-assignment operator, that would fail to compile.

2
votes

template<typename T2> void operator = (T2&) is not the copy-assignment operator.

so the default one void operator = (const Sample&) is generated.

but your is a better match for:

Sample<int> obj1(2);
Sample<int> obj2(3);
obj2 = obj1;

But with

const Sample<int> obj1(2);
Sample<int> obj2(3);
obj2 = obj1;

You will have your expected result:

Demo