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

Question

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?

score 74 · Accepted Answer · edited Dec 25 '12 at 04:28

74

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.

edited Dec 25 '12 at 04:28

Waleed Khan

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answered Dec 22 '12 at 13:49

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1

I never would have guessed. Thanks. – template boy Dec 22 '12 at 13:56
6

@templateboy: The comma operator always springs up in the most unexpected places ;) – Matthieu M. Dec 22 '12 at 14:38
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OMG we seriously need language constructs to not have to write such hacky tricks! O__O;;; – Klaim Dec 22 '12 at 16:54
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@Klaim: This is a lot nicer than SFINAE usually ended up being before C++11. – Ben Voigt Dec 22 '12 at 17:36
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@BenVoigt I agree, but this is more obscure! – Klaim Dec 22 '12 at 18:31
@Waleed Khan Will C++14 concepts end with SFINAE? – Alessandro Stamatto Feb 26 '13 at 17:52
Just curious if the use of "void()" itself is by at all possible just because we're in an unevaluated context – abigagli Aug 18 '14 at 16:41
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@abigagli Yes, `void` is an incomplete type so you can't call its "constructor" anywhere outside an unevaluated context like `decltype()` or `sizeof()`. – David G Oct 30 '14 at 01:17
@0x499602D2 Not true. You can use `void()` anywhere a void prvalue is allowed (granted, there's not very many...) – T.C. Mar 03 '15 at 20:09

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

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