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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?

like image 835
template boy Avatar asked Dec 22 '12 13:12

template boy


1 Answers

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.

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Pubby Avatar answered Sep 18 '22 13:09

Pubby