使用移动语义：右值引用作为方法参数

Question

I would like to doublecheck my understanding of move semantics. Am I missing anything in my reasoning:

#include <iostream>

using std::cout;

struct A
{
    A() {cout<<"Constructor\n";}
    ~A() {cout<<"Desctuctor\n";}
    A(const A&) {cout<<"Copy Constructor\n";}
    A(A&&) noexcept {cout<<"Move Constructor\n";}
    int x{1}; 
};

int f1(A a)
{
    cout<<"f1(A a)\n";
    return a.x;
}

int f2 (A&& a)
{
    cout<<"f2 (A&& a)\n";
    return a.x;
}

int main()
{
  A a1, a2;
  f1(std::move(a1));
  f2(std::move(a2));
}

Output:

Constructor
Constructor
Move Constructor
f1(A a)
Desctuctor
f2 (A&& a)
Desctuctor
Desctuctor

From what I can see with f2() , I'm not creating any additional copies, even "light" move copies, which is cool. So my understanding now is that if I'm going to use move semantics, I better always write functions/methods signature to accept r-values as to avoid any unnecessary copies at all. Or there is something else I'm missing?

Answer 1

You're almost always best just taking inputs you're going to keep by value (except for odd types which have very high costs of moves).

It maintains the same level of performance (in common scenarios) and keeps your code very simple to understand. It also makes it very clear that your code is keeping the value (because it's guaranteed to get moved out of), as an rvalue reference doesn't enforce that.

As you can see, the timings and generated code are identical for your examples when the print statements are removed: http://quick-bench.com/ADU4fzd0ISk0UrLboVhC-Pd7RmI