Re: Automatically create initializer lists

From:
=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 22 Nov 2012 11:31:59 -0800 (PST)
Message-ID:
<k8lsqo$9g2$1@dont-email.me>
Am 22.11.2012 06:28, schrieb fmatthew5876:

Is it possible to create a variable length initializer list from an
constant expression and fill the elements of the list with some
pattern?

Basically what I want to do is this:

struct Vec2 {
Vec2(float x, float y);
enum { size = 2; }
}

struct Vec3 {
Vec3(float x, float y, float z);
enum { size = 3; }
}

template <typename T> foo() {
T v SOME_COMPILE_TIME_MAGIC(T::size);
}

Where SOME_COMPILE_TIME_MAGIC(2) = {0, 1}
and SOME_COMPILE_TIME_MAGIC(3) = {0, 1, 2}

Basically calling the T() constructor for 2 or 3 parameters as needed.


OK, the following approach does *literally* what you are asking for,
but I'm cheating in one detail:

//------------
template<int...> struct indices {};

template<int I, class IndexTuple, int N>
struct make_indices_impl;

template<int I, int... Indices, int N>
struct make_indices_impl<I, indices<Indices...>, N>
{
    typedef typename make_indices_impl<I + 1, indices<Indices..., I>,
      N>::type type;
};

template<int N, int... Indices>
struct make_indices_impl<N, indices<Indices...>, N>
{
    typedef indices<Indices...> type;
};

template<int N>
struct make_indices : make_indices_impl<0, indices<>, N> {};

template<typename T>
struct maker
{
    template<int... I>
    constexpr static T _(indices<I...>)
    {
      return T{I...};
    }
};

template<typename T, int N>
constexpr T make_braced()
{
    return maker<T>::_(typename make_indices<N>::type{});
}

#define SOME_COMPILE_TIME_MAGIC(N) = make_braced<decltype(v), N>()
//------------

My more serious recommendation would change your code to

template <typename T> void foo()
{
   T v = make_braced<T, T::size>();
}

HTH & Greetings from Bremen,

Daniel Kr?gler

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