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When to return by value


When to Return by Value from a Function?

If you've been following my previous blogs, you might be eager to use std::move. At some point, you might consider using it to return a value from a function. This blog will discuss whether that's a good idea or not.

Proper Use Case for Returning by Move

Imagine you have a function that receives an r-value reference and performs some operation on it, then wants to return it. In this scenario, using std::move could be advantageous.

Tensor operator+(Tensor&& lhs, const Tensor& rhs){
    lhs += rhs;
    return std::move(lhs);

Since lhs is an l-value (because it's a parameter) bound to an r-value reference, using std::move is beneficial. If Tensor doesn't support move construction, that's fine too because in that case, the copy constructor will be called.

Proper Use Case for Returning by Forward

Let's adjust our previous example to work with universal references.

template <typename T>
Tensor sum(T&& tensor){
    return std::forward<T>(tensor); // moves r-value; copies l-value

In this function, the return statement acts as std::move for r-values and a no-op for l-values. This means the move constructor will be called for r-values and the copy constructor for l-values.

When Not to Use Move and Forward

In the C++ standard, there are two rules stating that compilers may elide the copying or moving of objects that are returned by value if:

  1. The type of the local object is the same as that returned by the function.
  2. The local object is what's being returned.

Let's look at an example:

Tensor makeTensor(){
    Tensor t;
    return t;

This copies t to the returned Tensor, but in reality, the compiler will perform return value optimization (RVO) and elide the copy because the first two conditions were met. However, if you use std::move in the return, you will hurt performance because now there won't be any RVO since std::move produces an r-value reference and doesn't satisfy the conditions mentioned before. So it will force the compiler to not perform RVO and perform a move or copy.

Let's say you are insistent on using std::move because you don't trust the compiler. For you, there is another rule: 3. If compilers choose not to perform copy elision, the object being returned must be treated as an r-value.

So, for our makeTensor function, if the compiler chooses not to perform RVO, it has to cast the return value to an r-value. In that case, makeTensor will look like this:

Tensor makeTensor(){
    Tensor t;
    return std::move(t);

All that to say:

Do not over-optimize your code. The compiler is mostly one step ahead.

I hope you enjoyed this blog.