Condition in Template Definition

Question

I want to change the return type of a template function depending on a property of the given type. Is there a possibility to do something like this, maybe with partial specialization (one for the cool T and one for not cool ones)?

template<typename T, typename ret = T::IsCool ? int : float>
inline ret get() {}

(It is always guaranteed, that T has the bool property IsCool.)

@nwp You do know that you're supposed to put answers in the answer box below, right? — Rakete1111, Jun 02 '17 at 12:40
@Rakete1111 I was under the impression that link-only answers are not answers, but feel free to add some text and make it one. — nwp, Jun 02 '17 at 12:41
@nwp Yes, but explaining it a bit and providing an example makes a good answer :) — Rakete1111, Jun 02 '17 at 12:41
Do note that if you want to change the return based on whether `IsCool` returns true or false then it needs to be `constexpr`. — NathanOliver, Jun 02 '17 at 12:44
In my real project it defines if `T` is allowed to appear multiple times. Therefore I want to decide to return just a `T*` or a `std::vector&`. — PaulProgrammerNoob, Jun 02 '17 at 12:44

score 8 · Accepted Answer · edited Jun 02 '17 at 12:56

8

You can use std::conditional to achieve this.

See http://en.cppreference.com/w/cpp/types/conditional

You can use it like so:

C++11

template<typename T, typename ret = std::conditional<T::IsCool, int, float>::type>
inline ret get() {}

C++14

template<typename T, typename ret = std::conditional_t<T::IsCool, int, float>>
inline ret get() {}

edited Jun 02 '17 at 12:56

NathanOliver

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answered Jun 02 '17 at 12:42

Bathsheba

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Condition in Template Definition

1 Answers1