decltype on type expressions

Question

Is there any way to avoid the dummy functions in the following example?

template<class T1, class T2>
struct A {

    static T1 T1_ ();
    static T2 T2_ ();

    typedef decltype (T1_ () + T2_ ()) sum_type;
};

I would like to write

typedef decltype (T1+T2) sum_type;

but that's not possible since T1 and T2 are types, not variables. Is my above solution really the easiest one possible?

typedef decltype(*(T1*)0 + *(T2*)0) sum_type; avoids the functions. — Vaughn Cato
– Vaughn Cato, Commented Nov 17, 2012 at 6:30

Cheers and hth. - Alf · Accepted Answer · 2012-11-17 06:38:43Z

5

The Holy Standard provides std::declval for exactly this purpose:

typedef decltype (declval<T1>()+declval<T2>()) sum_type;

Include the <utility> header.

answered Nov 17, 2012 at 6:38

Cheers and hth. - Alf

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3 Comments

JohnB Over a year ago

Thanks! Exactly what I need. Never heard about it. Visual C++ 2010 does not seem to have it, but now declared it myself by template<class T> T declval ();.

Xeo Over a year ago

@JohnB: The correct one is template<class T> typename std::add_rvalue_reference<T>::type declval();.

JohnB Over a year ago

Hm, shouldn't be the result of a function an rvalue anyway? I fully trust in what you say, but I do not understand why simple T does not suffice.

Vaughn Cato · Accepted Answer · 2012-11-17 06:38:35Z

4

You can do this:

typedef decltype(*(T1*)0 + *(T2*)0) sum_type;

answered Nov 17, 2012 at 6:38

Vaughn Cato

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