C++ constexpr at compile time

问题

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;
}

回答1:

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.

来源：https://stackoverflow.com/questions/12785691/c-constexpr-at-compile-time