Repeating a list

[Wesley Smith]

I'm trying to figure out how to take a list list<0, 1, 2> and turn it into a list list<0, 0, 0, 1, 1, 1, 2, 2, 2>.

I've tried various combinations of:

flatten<transform<List, filled_list<_1, 3>>>

The above code gives an error about _1 not having a type member, so I made a lazy version of flatten:

namespace lazy
{
    template <typename Sequence>
    using flatten = typename detail::flatten_impl<Sequence>;
}

and then did

transform<List, lazy::filled_list<_1, 3>>

Now the problem is that I have a list of filled_list_impl types, and I can't figure out how to unwrap them to flatten them. Thoughts?

[Edouard A.]

Could you share the code?

[Wesley Smith]

using List = brigand::list<
	brigand::uint16_t<0>,
	brigand::uint16_t<1>,
	brigand::uint16_t<2>>;

using Expanded = brigand::transform<List, brigand::filled_list<brigand::_1, 3>>;

int main(int argc, const char * argv[])
{
	brigand::for_each<Expanded>([](auto v)
	{
		using T = brigand::type_from<decltype(v)>;
		std::cout << typeid(T).name() << "\n";
	});
	return 0;
}

[Wesley Smith]

Using the lazy version, this one compiles:

using List = brigand::list<
	brigand::uint16_t<0>,
	brigand::uint16_t<1>,
	brigand::uint16_t<2>>;

using Expanded = brigand::transform<List, brigand::lazy::filled_list<brigand::_1, 3>>;

int main(int argc, const char * argv[])
{
	brigand::for_each<Expanded>([](auto v)
	{
		using T = brigand::type_from<decltype(v)>;
		std::cout << typeid(T).name() << "\n";
	});
	return 0;
}

But each element isn't a sequence, so flatten won't give me a 9 element list.

[Wesley Smith]

I don't have a solution for this exact problem right now, but my understanding of the lambda system (bind, apply, defer) has improved a lot after reading seeing the cartesian product examples and the discussion here: #187 . This is enough for me to close this out.