Given I have that type
template<int ...Is>
struct A {};
Can I "generate" the type A<0, 1, 2, 3, 4, 5,..., d> just from an integer d?
I thought about something like
template<int d>
struct B : A<std::index_sequence<d>...> {}
but it doesn't work.
Other option is to specialize manually:
template<int d>
struct B;
template<>
struct B<0>: A<> {};
template<>
struct B<1>: A<0> {};
template<>
struct B<2>: A<0, 1> {};
template<>
struct B<3>: A<0, 1, 2> {};
but obviously I don't be able to write B<3000> b;
[edit] my actual use-case is a "bit" more complex than that. I don't want to reimplement std::integer_sequence, but something more complex.
We already have what you want in the Standard library - std::make_integer_sequence. If you want to use your own type A<...> you can do this:
template<int... Is>
struct A {};
template<class>
struct make_A_impl;
template<int... Is>
struct make_A_impl<std::integer_sequence<int, Is...>> {
using Type = A<Is...>;
};
template<int size>
using make_A = typename make_A_impl<std::make_integer_sequence<int, size>>::Type;
And then for A<0, ..., 2999> write
make_A<3000>