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Home/ Questions/Q 8648651
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T13:21:45+00:00 2026-06-12T13:21:45+00:00

Am I right to think that this function should only be evaluated at compile

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Am I right to think that this function should only be evaluated at compile time, or is there a run-time cost to it?

template <typename T>
size_t constexpr CompID() {
    return typeid(T).hash_code();
}

struct Foo {};

int main(int argc, const char * argv[]) {
    size_t foo = CompID<Foo>();
    return 0;
}
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  1. Editorial Team
    Editorial Team
    2026-06-12T13:21:46+00:00Added an answer on June 12, 2026 at 1:21 pm

    constexpr function allows the function to be evaluated at compile time, but does not require that, so your answer is “maybe”. It depends on the compiler’s optimization settings.

    §7.1.5[dcl.constexpr]/7

    A call to a constexpr function produces the same result as a call to an equivalent non-constexpr function
    in all respects except that a call to a constexpr function can appear in a constant expression.

    If you wish to have no runtime cost, you could force compile-time evaluation by assigning it to a constexpr variable, e.g.

    constexpr auto foo = CompID<Foo>();
    

    Also note that type_info.hash_code() cannot be evaluated in compile-time (it is not a constexpr function, §18.7.1[type.info]/7). So your code is actually wrong.

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