User defined literal arguments are not constexpr?
- by Pubby
I'm testing out user defined literals. I want to make _fac return the factorial of the number.
Having it call a constexpr function works, however it doesn't let me do it with templates as the compiler complains that the arguments are not and cannot be constexpr.
I'm confused by this - aren't literals constant expressions? The 5 in 5_fac is always a literal that can be evaluated during compile time, so why can't I use it as such?
First method:
constexpr int factorial_function(int x) {
return (x > 0) ? x * factorial_function(x - 1) : 1;
}
constexpr int operator "" _fac(unsigned long long x) {
return factorial_function(x); // this works
}
Second method:
template <int N> struct factorial {
static const unsigned int value = N * factorial<N - 1>::value;
};
template <> struct factorial<0> {
static const unsigned int value = 1;
};
constexpr int operator "" _fac(unsigned long long x) {
return factorial_template<x>::value; // doesn't work - x is not a constexpr
}
