How should this c++ typedef using decltype and declval be written to make it portable?

Question

I have the following

template <typename F, typename A0>
struct ResultOf {
        typedef typename decltype(boost::declval<F>()(boost::declval<A0>())) Type;
};

It was written so that VS2010 could have a result_of that worked for a specific use case. It is working under vs2015, vs2013 and vs2010 but under gcc I get a compile error

error: expected nested-name-specifier before ‘decltype’
typedef typename decltype(boost::declval<F>()(boost::declval<A0>())) Type;

Is there an obvious small fix here?

How "portable" does it need to be? old VS is a totally different beast, hardly fit to be called C++ — xaxxon, Oct 16 '18 at 07:51

score 2 · Accepted Answer · answered Oct 16 '18 at 07:53

2

typename keyword is not needed here. It is used, in particular, to denote a dependent type, like T::value_type, when a compiler cannot know whether value_type is a type. There are no dependent types in the present case.

answered Oct 16 '18 at 07:53

Evg

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score 2 · Answer 2 · answered Oct 16 '18 at 07:55

2

Removing the typename works in clang, gcc, and modern msvc:

https://godbolt.org/z/CfOw-_

answered Oct 16 '18 at 07:55

xaxxon

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How should this c++ typedef using decltype and declval be written to make it portable?

2 Answers2