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; } 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
constexprfunction produces the same result as a call to an equivalent non-constexprfunction in all respects except that a call to aconstexprfunction 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.
constexpr means "I can do anything at compile-time". It's very limited, on purpose.type2int struct in stackoverflow.com/a/1708628/224671 for example.