Should C++ std::future<T> method be named is_ready() or ready()?

Question

Why is the experimental name is_ready() considered as an enhancement to std::future and not ready() which is more consistent with the STL coding style? Future already has a method called valid() which also doesn't have the is_ prefix.

Interestingly both N3721 and N3865 contain both names.

Answer 1

I suspect it's because of already existing enum constant std::future_status::ready. So is_ready() checks for the status ready. Although they both are in different naming scopes, I assume authors wish to avoid name intersection.

Answer 2

<speculation>

I suspect is_ready() is clearer. When reading the code if (future.is_ready()) almost sound english and obviously means that is_ready is const. if ( future.ready() ) sounds like it could mean "future needs to get ready, then return" or "future is already ready".

</speculation>