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Overloading on R-value references and code duplication

Asked 2011-05-15T04:39:46.157
17

Consider the following:

struct vec
{
    int v[3];

    vec() : v() {};
    vec(int x, int y, int z) : v{x,y,z} {};
    vec(const vec& that) = default;
    vec& operator=(const vec& that) = default;
    ~vec() = default;

    vec& operator+=(const vec& that)
    {
        v[0] += that.v[0];
        v[1] += that.v[1];
        v[2] += that.v[2];
        return *this;
    }
};

vec operator+(const vec& lhs, const vec& rhs)
{
    return vec(lhs.v[0] + rhs.v[0], lhs.v[1] + rhs.v[1], lhs.v[2] + rhs.v[2]);
}
vec&& operator+(vec&& lhs, const vec& rhs)
{
    return move(lhs += rhs);
}
vec&& operator+(const vec& lhs, vec&& rhs)
{
    return move(rhs += lhs);
}
vec&& operator+(vec&& lhs, vec&& rhs)
{
    return move(lhs += rhs);
}

Thanks to r-value references, with these four overloads of operator+ I can minimize the number of objects created, by reusing temporaries. But I don't like the duplication of code this introduces. Can I achieve the same with less repetition?

c++ rvalue-reference c++11
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1 Answer

9

Since your vec type is "flat" (there is no external data), moving and copying do exactly the same thing. So all your rvalue references and std::moves gain you absoutely nothing in performance.

I would get rid of all additional overloads and just write the classic reference-to-const version:

vec operator+(const vec& lhs, const vec& rhs)
{
    return vec(lhs.v[0] + rhs.v[0], lhs.v[1] + rhs.v[1], lhs.v[2] + rhs.v[2]);
}

In case you have little understanding of move semantics yet, I recommend studying this question.

Thanks to r-value references, with these four overloads of operator+ I can minimize the number of objects created, by reusing temporaries.

With a few exceptions, returning rvalue references is a very bad idea, because calls of such functions are xvalues instead of prvalues, and you can get nasty temporary object lifetime problems. Don't do it.

answered 2011-05-15T11:37:59.173

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