What is the advantage of && in this code?

Question

In the following code, what is the advantage of using &&? Code from the answer for Specialize the same operator for different properties

From this question, I understand that an argument &&means that it is a link that can be changed by a function.

decay_tprobably does not allow the compiler to interpret the reference to the variable as an array, as in What is std :: decay and when should it be used?

std::forward- perfect forwarding as described here . Why do we need this shipment?

Thank.

#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 1

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X:: operator = - X X, X &, const X &, & Const volatileX &.

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Bb = static_cast<const A&>(Aa);

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@Jarod42:

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