Is it possible to construct the elements of a member array depending on an integral template parameter?

Question

Assume:

template<class T,int N>
struct A {
  A(): /* here */ {}

  T F[N];
};

I need the elements of F[] to be constructed with {0,1,2,...,N-1}. If possible I would like to avoid recursively defined template structs with defining the last level as template<class T> struct A<T,0> and doing some complicated template tricks. Can C++11 initializer lists help?

This is similar Template array initialization with a list of values, but it does not construct the elements with the increasing value. It sets it later in a run-time loop.

why can't array be used here ? – Jagannath Oct 11 '12 at 09:59 — Jagannath, Oct 11 '12 at 09:59
It can, but doesn't solve this issue – ritter Oct 11 '12 at 09:59 — ritter, Oct 11 '12 at 09:59

score 4 · Answer 1 · edited May 23 '17 at 11:47

4

You can do this with a variadic value template and constructor delegation:

template<int... I> struct index {
    template<int n> using append = index<I..., n>; };
template<int N> struct make_index { typedef typename
    make_index<N - 1>::type::template append<N - 1> type; };
template<> struct make_index<0> { typedef index<> type; };
template<int N> using indexer = typename make_index<N>::type;

template<class T, int N>
struct A {
  template<T...I> A(index<I...>): F{I...} {}

  A(): A(indexer<N>{}) {}

  T F[N];
};

This uses the sequence pack generator from Calling a function for each variadic template argument and an array

edited May 23 '17 at 11:47

Community

1
1

answered Oct 11 '12 at 10:05

ecatmur

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Is the use of `using` essential to the solution? – ritter Oct 11 '12 at 10:12
@Frank no, it can always be replaced by `typedef`. It's just convenient in this case. – ecatmur Oct 11 '12 at 10:43
I voted up - I love how templates can be completely unreadable in the new standard ;-) – Pawel Zubrycki Oct 11 '12 at 13:32
1

@Pawel: A bit of formatting would've helped a ton, and maybe even [another implementation of the indices trick](http://loungecpp.wikidot.com/tips-and-tricks:indices). – Xeo Oct 11 '12 at 18:25

score 2 · Answer 2 · edited May 23 '17 at 12:18

2

Assuming some kind of indices solution is available:

A(): A(make_indices<N>()) {}

// really a private constructor
template<int... Indices>
explicit A(indices<Indices...>)
    // Can be an arbitrary expression or computation, too, like
    // (Indices + 3)...
    : F {{ Indices... }}
{}

If your compiler doesn't support delegating constructors, one option is to switch to std::array<T, N> and use a private static helper that returns an initialized array, such that the default constructor would become:

A(): F(helper(make_indices<N>())) {}

This would of course incur an additional (move) construction.

edited May 23 '17 at 12:18

Community

1
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answered Oct 11 '12 at 09:55

Luc Danton

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[Complete demo on LWS](http://liveworkspace.org/code/9894e8cb01d5f7af1bbbef5e04397a16). – Luc Danton Oct 11 '12 at 10:00
Is this `using` entirely necessary. My compiler GCC 4.6 doesn't understand it – ritter Oct 11 '12 at 10:09
@Frank You can introduce type synonyms with `typedef` as well. – Luc Danton Oct 11 '12 at 10:14
If I do so, I get: error: type ‘A’ is not a direct base of ‘A’ – ritter Oct 11 '12 at 10:18
@Frank 4.6 doesn't support delegating constructors. – Luc Danton Oct 11 '12 at 10:18
Last question: What kind of thing is `return {};` ? Never seen this before. Does it return whatever is the return type by calling the default constructor? – ritter Oct 11 '12 at 10:30
@Frank Same braces as in `T t = {};`. Try to search for 'uniform initialization'. – Luc Danton Oct 11 '12 at 10:31

Is it possible to construct the elements of a member array depending on an integral template parameter?

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