I want to do the following:
// have a constexpr function
template<class T>
constexpr T square( T const i )
{
return i * i;
}
// transform a std::integer_sequence<> by calling the constexpr function on every integer
template<class Fn, class T, T... values>
static constexpr auto make_type( Fn fn, std::integer_sequence<T, values...> )
{
return std::integer_sequence<T, fn( values )...>{};
}
// so that I can use it like so
using type = decltype( make_type( square, std::integer_sequence<int, 1, 2, 3>{} ) );
However, I get the following error:
...\main.cpp|19|error: 'fn' is not a constant expression|
fn is not usable in a constant expression - it's a bog-standard block-scope variable. You have to pass the functor as a type.
template <typename Fn, typename T, T... values>
static constexpr std::integer_sequence<T, Fn{}(values)...>
make_type(std::integer_sequence<T, values...>) {return {};}
And rewrite your function as
struct Square {
template <typename T> constexpr T operator()(T const& t)
{return t*t;}
};
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