How to move from std::optional<T>

Question

Consider the following example where we parse data and pass the result to the next function:

Content Parse(const std::string& data); void Process(Content content);  int main() {     auto data = ReadData();     Process(Parse(data));     } 

Now let's change the code using std::optional to handle a failed parsing step:

optional<Content> Parse(const std::string& data); void Process(Content content);  int main() {     auto data = ReadData();     auto content = Parse(data);     if (content)         Process(move(*content)); } 

Is it valid to move from optional<T>::value()? If it's ok for std::optional is it valid for boost::optional as well?

Answer 1

It is valid to move from optional<T>::value() since it returns a mutable reference and the move does not destroy the object. If the optional instance is not engaged, value() will throw a bad_optional_access exception (§20.6.4.5).

You explicitly check whether the option is engaged:

if (content)     Process(move(*content)); 

But you don't use the member value() to access the underlying T. Note that value() performs a check internally before returning a valid T&, unlike operator* which has a precondition that the optional instance shall be engaged. This is a subtle difference, but you use the right idiom:

if (o)   f(*o) 

as opposed to

if (o)  // redundant check   f(o.value()) 

In Boost, the situation is a little different: first, there exists no member function called value() that provides checked access. (A bad_optional_access exception simply does not exist). The member get() is just an alias for operator* and always relies on the user checking that the optional instance is engaged.