Suppose the following function
template<size_t N>
constexpr std::array<float, N> make_ones()
{
std::array<float, N> ret{};
for (size_t k = 0; k != N; ++k)
{
ret[k] = 1.0f;
}
return ret;
}
Is it possible to write that with a fold expression? The problem is that I do not have a pack to expand.
Is it possible to write that with a fold expression?
Not with fold expression but pack expansion as well as with index sequence, in c++20 you could do:
template <size_t N>
constexpr std::array<float, N> make_ones()
{
return []<size_t... Is>(std::index_sequence<Is...>) {
return std::array<float, sizeof...(Is)>{( static_cast<void>(Is), 1.0f)...};
}(std::make_index_sequence<N>{});
}
For compilers that doesn't support C++20 or later, you might do
template <size_t... Is>
constexpr std::array<float, sizeof...(Is)> make_ones_impl(std::index_sequence<Is...>)
{
return std::array<float, sizeof...(Is)>{(static_cast<void>(Is), 1.0f)...};
}
template<size_t N>
constexpr std::array<float, N> make_ones()
{
return make_ones_impl(std::make_index_sequence<N>{});
}
std::fillconstexpr->constevalto be absolutely certain that the array is filled at compile time (else compiler error will ensue).