What does the "template" keyword mean when places after scope resolution operator (::)

Asked Jan 29 '16 at 19:00

Active Jan 29 '16 at 19:24

Viewed 123 times

I came accross the following construct while reading this article:

template<typename F, typename...As>
using meta_apply = typename F::template apply<As...>;

Can someone explain to me what does the template keyword mean when it's placed after the scope resolution operator ::? Or better point me to where this usage is described in C++ reference of any kind. I tried looking it up myself but couldn't find any.

EDIT

Consider also this code from the same article:

meta_compose<F0, meta_compose<F1, F2>::template apply>::template apply

Here, the last apply token is not followed either < or > thus its usage is not explained in either of the two alternative questions suggested.

edited Jan 29 '16 at 19:24

asked Jan 29 '16 at 19:00

Levi Haskell

1

See also: http://stackoverflow.com/questions/610245/where-and-why-do-i-have-to-put-the-template-and-typename-keywords. – GManNickG Jan 29 '16 at 19:04
I don't think it's a duplicate, please see my edit – Levi Haskell Jan 29 '16 at 19:31
2

It tells the compiler that the thing afterwards names a template, so that 1) it knows how to parse ` – T.C. Jan 29 '16 at 19:40
In other words to be able to bind it `template class F0` template argument above (in the article)? - I think you answered my question. – Levi Haskell Jan 29 '16 at 20:01

What does the "template" keyword mean when places after scope resolution operator (::)

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