When does a constexpr function get evaluated at compile time?

Question

Since it is possible that a function declared as constexpr can be called during run-time, under which criteria does the compiler decide whether to compute it at compile-time or during runtime?

template<typename base_t, typename expo_t>
constexpr base_t POW(base_t base, expo_t expo)
{
    return (expo != 0 )? base * POW(base, expo -1) : 1;
}

int main(int argc, char** argv)
{
    int i = 0;
    std::cin >> i;

    std::cout << POW(i, 2) << std::endl;
    return 0;
}

In this case, i is unknown at compile-time, which is probably the reason why the compiler treats POW() as a regular function which is called at runtime. This dynamic however, as convenient as it may appear to be, has some impractical implications. For instance, could there be a case where I would like the compiler to compute a constexpr function during compile-time, where the compiler decides to treat it as a normal function instead, when it would have worked during compile-time as well? Are there any known common pitfalls?

Answer 1

constexpr functions will be evaluated at compile time when all its arguments are constant expressions and the result is used in a constant expression as well. A constant expression could be a literal (like 42), a non-type template argument (like N in template<class T, size_t N> class array;), an enum element declaration (like Blue in enum Color { Red, Blue, Green };, another variable declared constexpr, and so on.

They might be evaluated when all its arguments are constant expressions and the result is not used in a constant expression, but that is up to the implementation.