User defined literal arguments are not constexpr?
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Published on 2011-11-12T23:52:25Z
Indexed on
2011/11/13
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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
}
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