Specialize same operator for different traits

Question

I want to do the following with specialization by traits.

1. Array Aa = Scalar in_a would use overload I.

2. Array Aa = Array Bb would use overload II.

In the following code, overload II never get used.

Someone mentioned that T1 cannot be deduced in overload II.

How to fix that?

I used the C++ shell to compile the code with C++14.

#include <iostream>
#include <type_traits>

using namespace std;
class A; // forward declaration.

template <typename T>
struct is_A : false_type {};
template <> struct is_A<A> : true_type {};

template <typename T>
struct is_int : false_type {};
template <> struct is_int<int> : true_type {};
template <> struct is_int<long> : true_type {};

class A{
    public:
        int val;
        void print(void){
            std::cout << val << std::endl;
        }
        template <typename T1>
        enable_if_t<is_int<T1>::value,void>
        operator=(const T1 & input){
            val = 2*input; //Overload I
        }
        template <typename T1>
        enable_if_t<is_A<T1>::value,void>
        operator=(const T1 & Bb){
            val = 5*Bb.val; //Overload II
        }
};

int main(void){
    A Aa;
    A Bb;
    int in_a = 3;
    Aa = in_a; //This uses overload I as intended.
    Bb = Aa; //I want this to use overload II, but
             //actually overload I is used.
             //This leads to an error during compilation.
    Aa.print(); //This should give 6. (3x2)
    Bb.print(); //This should give 30. (6x5)
}

Answer 1

Your code should be

template <typename T>
std::enable_if_t<is_int<T>::value, A&>
operator=(const T& input){
    val = 2 * input; //Overload I
    return *this;
}
template <typename T>
std::enable_if_t<is_A<T>::value, A&>
operator=(T& rhs){
    val = 5 * rhs.val; //Overload II
    return *this;
}

Demo

But even simpler in your case

A& operator=(int input){
    val = 2 * input; //Overload I
    return *this;
}

A& operator=(const A& rhs){
    val = 5 * rhs.val; //Overload II
    return *this;
}

Answer 2

Here is your code simplified and working as intended:

#include <iostream>
#include <type_traits>
#include<utility>

class A;

template <typename T>
struct is_A : std::false_type {};
template <> struct is_A<A> : std::true_type {};

template <typename T>
struct is_int : std::false_type {};
template <> struct is_int<int> : std::true_type {};
template <> struct is_int<long> : std::true_type {};

class A{
public:
    int val;

    void print(void){
        std::cout << val << std::endl;
    }

    template <typename T1>
    std::enable_if_t<is_int<std::decay_t<T1>>::value, void>
    operator=(T1 && input){
        val = 2*std::forward<T1>(input);
    }

    template <typename T1>
    std::enable_if_t<is_A<std::decay_t<T1>>::value,void>
    operator=(T1 && Bb){
        val = 5*std::forward<T1>(Bb).val;
    }
};

int main(void){
    A Aa;
    A Bb;
    int in_a = 3;
    Aa = in_a;
    Bb = Aa;
    Aa.print(); //This should give 6. (3x2)
    Bb.print(); //This should give 30. (6x5)
}

Answer 3

Do you really need all the template magic for your simple case?

#include <iostream>

class A;

class A{
public:
    int val;

    void print(void){
        std::cout << val << std::endl;
    }

    void operator =(const A& in){ val = in.val*5; }
    void operator =(int in) { val = in*2; }
};


int main(void){
    A Aa;
    A Bb;
    Aa = 3;
    Bb = Aa;
    Aa.print(); //This should give 6. (3x2)
    Bb.print(); //This should give 30. (6x5)
    return 0;
}