My follow-up question is: why does decltype((value)) return int& in C++?

You can find an explanation here,

https://www.scs.stanford.edu/~dm/blog/decltype.html

There are a lot of words on that page and none really jump out at me as explaining why C++ chose to have decltype yield something different for decltype(value) and decltype((value)). But I was unable to read them all, so maybe it is buried somewhere in there.

Sorry, could not find a video.

See decltype section.

It provides a series of rules for exactly when decltype() picks a reference or non-reference type, but as far as I can see, it does not explain why the language chose to sometimes return references in some scenarios, such as for '(value)' vs 'value'.

Because decltype(*&value) is int &.

The real question to ask is why decltype(value) is int.

That doesn't really answer the question. It's not obvious why decltype(*&value) is int& either.

Normally decltype(E) returns the type of the expression E. The type of the expressions *&value and value are both int& for instance. However they also wanted a way to query the type an identifier was declared with, eg int x would be int. For some reason they chose to expose this by overriding decltype(E) when E is a plain identifier. decltype((value)) avoids this special case and returns the type of the expression (value), which is obviously the same as the expression value.

Thank you, I looked specifically whether this was discussed here already and found your reference.