Questions about the move assignment operator

Question

Imagine the following class that manages a resource (my question is only about the move assignment operator):

struct A
{
    std::size_t s;
    int* p;
    A(std::size_t s) : s(s), p(new int[s]){}
    ~A(){delete [] p;}
    A(A const& other) : s(other.s), p(new int[other.s])
    {std::copy(other.p, other.p + s, this->p);}
    A(A&& other) : s(other.s), p(other.p)
    {other.s = 0; other.p = nullptr;}
    A& operator=(A const& other)
    {A temp = other; std::swap(*this, temp); return *this;}
    // Move assignment operator #1
    A& operator=(A&& other)
    {
        std::swap(this->s, other.s);
        std::swap(this->p, other.p);
        return *this;
    }
    // Move assignment operator #2
    A& operator=(A&& other)
    {
        delete [] p;
        s = other.s;
        p = other.p;
        other.s = 0;
        other.p = nullptr;
        return *this;
     } 
};

Question:

What are the advantages and disadvantages of the two move assignment operators #1 and #2 above? I believe the only difference I can see is that std::swap preserves the storage of the lhs, however, I don't see how that would be useful as rvalues would be destroyed anyways. Maybe the only time would be with something like a1 = std::move(a2);, but even in this case I don't see any reason to use #1.

Answer 1

This is a case where you should really measure.

And I'm looking at the OP's copy assignment operator and seeing inefficiency:

A& operator=(A const& other)
    {A temp = other; std::swap(*this, temp); return *this;}

What if *this and other have the same s?

It seems to me that a smarter copy assignment could avoid making a trip to the heap if s == other.s. All it would have to do is the copy:

A& operator=(A const& other)
{
    if (this != &other)
    {
        if (s != other.s)
        {
            delete [] p;
            p = nullptr;
            s = 0;
            p = new int[other.s];
            s = other.s;
        }
        std::copy(other.p, other.p + s, this->p);
    }
    return *this;
}

If you don't need strong exception safety, only basic exception safety on copy assignment (like std::string, std::vector, etc.), then there is a potential performance improvement with the above. How much? Measure.

I've coded this class three ways:

Design 1:

Use the above copy assignment operator and the OP's move assignment operator #1.

Design 2:

Use the above copy assignment operator and the OP's move assignment operator #2.

Design 3:

DeadMG's copy assignment operator for both copy and move assignment.

Here is the code I used to test:

#include <cstddef>
#include <algorithm>
#include <chrono>
#include <iostream>

struct A
{
    std::size_t s;
    int* p;
    A(std::size_t s) : s(s), p(new int[s]){}
    ~A(){delete [] p;}
    A(A const& other) : s(other.s), p(new int[other.s])
    {std::copy(other.p, other.p + s, this->p);}
    A(A&& other) : s(other.s), p(other.p)
    {other.s = 0; other.p = nullptr;}
    void swap(A& other)
    {std::swap(s, other.s); std::swap(p, other.p);}
#if DESIGN != 3
    A& operator=(A const& other)
    {
        if (this != &other)
        {
            if (s != other.s)
            {
                delete [] p;
                p = nullptr;
                s = 0;
                p = new int[other.s];
                s = other.s;
            }
            std::copy(other.p, other.p + s, this->p);
        }
        return *this;
    }
#endif
#if DESIGN == 1
    // Move assignment operator #1
    A& operator=(A&& other)
    {
        swap(other);
        return *this;
    }
#elif DESIGN == 2
    // Move assignment operator #2
    A& operator=(A&& other)
    {
        delete [] p;
        s = other.s;
        p = other.p;
        other.s = 0;
        other.p = nullptr;
        return *this;
     } 
#elif DESIGN == 3
    A& operator=(A other)
    {
        swap(other);
        return *this;
    }
#endif
};

int main()
{
    typedef std::chrono::high_resolution_clock Clock;
    typedef std::chrono::duration<float, std::nano> NS;
    A a1(10);
    A a2(10);
    auto t0 = Clock::now();
    a2 = a1;
    auto t1 = Clock::now();
    std::cout << "copy takes " << NS(t1-t0).count() << "ns\n";
    t0 = Clock::now();
    a2 = std::move(a1);
    t1 = Clock::now();
    std::cout << "move takes " << NS(t1-t0).count() << "ns\n";
}

Here is the output I got:

$ clang++ -std=c++11 -stdlib=libc++ -O3 -DDESIGN=1  test.cpp 
$ a.out
copy takes 55ns
move takes 44ns
$ a.out
copy takes 56ns
move takes 24ns
$ a.out
copy takes 53ns
move takes 25ns
$ clang++ -std=c++11 -stdlib=libc++ -O3 -DDESIGN=2  test.cpp 
$ a.out
copy takes 74ns
move takes 538ns
$ a.out
copy takes 59ns
move takes 491ns
$ a.out
copy takes 61ns
move takes 510ns
$ clang++ -std=c++11 -stdlib=libc++ -O3 -DDESIGN=3  test.cpp 
$ a.out
copy takes 666ns
move takes 304ns
$ a.out
copy takes 603ns
move takes 446ns
$ a.out
copy takes 619ns
move takes 317ns

DESIGN 1 looks pretty good to me.

Caveat: If the class has resources that need to be deallocated "quickly", such as mutex lock ownership or file open-state ownership, the design-2 move assignment operator could be better from a correctness point of view. But when the resource is simply memory, it is often advantageous to delay deallocating it as long as possible (as in the OP's use case).

Caveat 2: If you have other use cases you know to be important, measure them. You might come to different conclusions than I have here.

Note: I value performance over "DRY". All of the code here is going to be encapsulated within one class (struct A). Make struct A as good as you can. And if you do a sufficiently high quality job, then your clients of struct A (which may be yourself) won't be tempted to "RIA" (Reinvent It Again). I much prefer to repeat a little code within one class, rather than repeat the implementation of entire classes over and over again.

Answer 2

It's more valid to use #1 than #2 because if you use #2, you're violating DRY and duplicating your destructor logic. Secondly, consider the following assignment operator:

A& operator=(A other) {
    swap(*this, other);
    return *this;
}

This is both copy and move assignment operators for no duplicated code- an excellent form.