How do I use concepts in if-constexpr?

Question

How does one use concepts in if constexpr?

Given the example below, what would one give to if constexpr to return 1 in case T meets the requirements of integral and else 0?

template<typename T>
concept integral = std::is_integral_v<T>;

struct X{};

template<typename T>
constexpr auto a () {
    if constexpr (/* T is integral */) {
        return 1;
    } 
    else {
        return 0;
    }
}

int main () {
    return a<X>();
}

This particular function can be simplified to `return integral;`. — L. F., Jul 14 '19 at 11:49

score 31 · Answer 1 · edited Feb 11 '19 at 13:38

31

Concepts are named boolean predicates on template parameters, evaluated at compile time.

In a constexpr if statement, the value of the condition must be a contextually converted constant expression of type bool.

So in this case, usage is simple:

if constexpr ( integral<T> )

edited Feb 11 '19 at 13:38

Peter Mortensen

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answered Feb 07 '19 at 13:10

P.W

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Is this still valid in the latest draft of the Standard? Or do we need `requires integral` instead? – Vittorio Romeo Aug 15 '19 at 15:44

nnolte · Accepted Answer · 2019-02-26T17:09:01.640

22

It is sufficient to do:

if constexpr ( integral<T> )

since integral<T> is already testable as bool

edited Feb 26 '19 at 17:09

answered Feb 07 '19 at 12:36

nnolte

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1

Is this still valid in the latest draft of the Standard? Or do we need `requires integral` instead? – Vittorio Romeo Aug 15 '19 at 15:44

How do I use concepts in if-constexpr?

2 Answers2