Operator overloading, Rvalues, C++

Question

Consider a struct

template<typename T, size_t N>
struct Something {
    std::array<T,N> args;

    // Some constructors

};

Now let's overload = operator for Something<T>. In fact I can implement it two ways.

First Way

Something& operator=(const Something& rhs){
    // an implementation with copy semantics
}

Something& operator=(Something&& rhs) {
    // an implementation with move semantics
}

Second Way

Something& operator=(Something rhs){
    // implement with move semantics
}

So, my question is what is the most standard way and the most optimal way to do overloading First Way or Second Way?

Are you asking in general, or for you particular case? Generally, the second way may not be optimal. — Daniel Langr, Dec 16 '20 at 06:47
It depend on you purpose , the choice of pointer refer and memory copy — ElapsedSoul, Dec 16 '20 at 06:51

score 2 · Accepted Answer · answered Dec 16 '20 at 06:54

2

For this particular case you should not implement the assignment operator. The compiler does that already for you:

#include <array>
#include <cstddef>

template<typename T, size_t N>
struct Something {
    std::array<T,N> args;
};

int main() {
    Something<int,42> a;
    Something<int,42> b;
    a = b;
}

Demo

For the general case I refer you to What are the basic rules and idioms for operator overloading?. And consider that not everything can be moved and not everything can be copied. Moreover, sometimes a move is just a copy. Hence, it depends.

answered Dec 16 '20 at 06:54

463035818_is_not_an_ai

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actually I was asking in general) – Hrant Nurijanyan Dec 16 '20 at 07:08
@HrantNurijanyan you wanted to ask for the general case, but you did ask for this example ;) – 463035818_is_not_an_ai Dec 16 '20 at 07:41
Couldn't you use `std::array::size_type` and (then) don't need to include cstddef? – JHBonarius Dec 16 '20 at 13:29

Operator overloading, Rvalues, C++

1 Answers1