Take value out of std::optional

Question

How do you actually take a value out of optional? Meaning take ownership of the value inside the std::optional and replace it with std::nullopt (or swap it with another value)?

In Rust for example you could .unwrap your Option or do something like foo.take().unwrap(). I'm trying to do something like that with C++ optionals.

Answer 1

operator*/value() returns a reference to the value held by the optional, so you can simply use std::move to move it to a temporary variable

std::optional<std::string> opt = "abc";
// "take" the contained value by calling operator* on a rvalue to optional
auto taken = *std::move(opt);

This will invoke the rvalue reference overload of operator*() of the optional, which returns an rvalue reference to the contained value.

You can also directly perform std::move on the return value of the operator*() of the lvalue optional, which will convert the lvalue reference of the contained value into an rvalue

auto taken = std::move(*opt);

Answer 2

The standard way in C++ to replace a value and return the old, like mem::replace in Rust, is std::exchange.

So:

std::exchange(o, std::nullopt).value();

https://godbolt.org/z/G7faK9rxj