Why use std::swap instead of std::move for (optional) out values?

Question

I am looking at code from Microsoft and I noticed the following:

An optional parameter std::vector<T>* vOut is passed into a function. The function itself uses a separate local std::vector<T> result for the optional result.

At the end of the function, they do the following to pass the result to the caller:

if(vOut) {
    std::swap(*vOut,result);
}

why would someone use a std::swap here instead of a move-assignment like *vOut = std::move(result)?

Here is an example

Answer 1

Well, there is always the possibility that something was made pre-C++11, so moves might not have been available.

However, I might occasionally use swap when I want the other object to be in a "better defined" state -- that is, move the data but also have a clear object, instead of just one that's merely safe to use.

And, last but not least, somebody might not think about every tiny detail -- not every piece of production code is meant as a textbook example, and std::move vs. std::swap, in many cases, is micro-optimization at best (the move might even be implemented on top of a swap member), so it's not something to be particularly concerned about.