Can the conditional operator ( ? : ) in C++ be compile time?

Question

Can the ternary (conditional) operator be used as an analogous to constexpr if(), introduced in C++17?

I would like to add some conditionality to member variables initialization in a template. Would the following expression resolve at compile time or runtime? If so, is there any other operator that resolves at compile time such that template specialisation can be avoided?

template<int a>
struct hello {
    constexpr static int n = (a != 0) ? 10 : 20;
}

"*such that template specialisation can be avoided*" Why do you want to avoid it? — Acorn, Jan 29 '20 at 17:24
You should probably use `constexpr` instead of `const`, especially if compile-time evaluation is important to you (although in this case it will most likely not make a difference.) — walnut, Jan 29 '20 at 17:31
If you only pass in compile-time constant expressions, any decent compiler should be able to evaluate it at compile time. — Jesper Juhl, Jan 29 '20 at 17:33

score 7 · Accepted Answer · answered Jan 29 '20 at 17:27

It depends on what you mean by "analogous to constexpr if()". if constexpr requires that the condition is a constant expression. It also has certain privileges in template code to discard the branches not taken.

?: does not have that functionality.

However ?: can appear in constant expressions just fine, and it always could. It doesn't make an expression non-constant.

score 2 · Answer 2 · answered Jan 29 '20 at 17:24

2

Yes, it absolutely can and in fact it could already be used in C++11 before the introduction of if constexpr, and even before C++11 in constant expressions, such as the one in your question.

answered Jan 29 '20 at 17:24

Konrad Rudolph

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Can the conditional operator ( ? : ) in C++ be compile time?

2 Answers2