CODEWITHSUNDEEP
c++c++11decltype
bjackfly
bjackfly

Reputation: 3336

Is there anyway to have a decltype on a template parameter?

quick question. If I have a function signature like 

template <typename T, typename ItType>
ItType binarySearch ( T mid, ItType first, ItType last );

Is there anyway to do something like the following? I know this syntax isn't correct but you get the idea as I can do a decltype similar with regular functions see below. The compiler knows the type of ItType at compile time so shouldn't it be able to deduce the type of *ItType as well?

template <typename ItType>
ItType binarySearch ( decltype(*ItType) mid, ItType first, ItType last );


// lambda
auto p = v.begin() + (v.end() - v.begin())/2;
std::partition ( v.begin(), v.end(), [p](decltype(*p) i) { return i < *p; } )

Upvotes: 3

Views: 1168

Answers (1)

Daniel Frey
Daniel Frey

Reputation: 56921

The problem with

decltype(*ItType)

is that *ItType is not a valid expression. A naive approach could look like this:

decltype(*ItType())

which would work if ItType is default constructible. Since you don't want to enforce that, you can use std::declval to "call" a function that pretends to return an instance of ItType:

decltype(*std::declval<ItType>())

This function is only declared but never defined which means you can not really call it, but that doesn't matter since you are using it within decltype(), which is an unevaluated context.

Upvotes: 1

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