CODEWITHSUNDEEP
c++c++11sfinaedecltypetrailing-return-type
template boy
template boy

Reputation: 10490

What does the 'void()' in 'auto f(params) -> decltype(..., void())' do?

I found code here that looked something like this:

auto f(T& t, size_t n) -> decltype(t.reserve(n), void()) { .. }

In all the documentation I read I was told that decltype is signed as:

decltype( entity )

or

decltype( expression )

And there is no second argument anywhere. At least that's what's pointed to on cppreference. Is this a second argument to decltype? And if so, what does it do?

Upvotes: 57

Views: 3563

Answers (1)

Pubby
Pubby

Reputation: 53097

Since it is an expression that comma is simply the comma operator (meaning the type is the type of the rhs side: void), not another argument.

That code is using SFINAE - it's enabled if t.reserve(n) exists but it wants to keep the return type as void.

Upvotes: 74

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