Copy constructor gets expression's chunk_shape if it is chunked

[davidbrochart]

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Description

If an expression is assigned to a chunked array without specifying a chunk shape, the chunk shape is taken from the expression if the expression is chunked, otherwise the chunk shape is the expression's shape, which basically means no chunk (or one chunk).
TODO: propagate chunk shape in expressions.

[davidbrochart]

It seems to work fine with gcc but not with clang.

[JohanMabille]

Maybe you can rely on void_t to detect expressions that provide the chunk_shape method? Something like:

template <class E, class = void>
struct get_chunk_shape
{
    static const auto& run(const xexpression<E>& e)
    {
        return e.derived_cast().shape();
    }
};

template <class E>
struct get_chunk_shape<E, void_t<decltype(std::declval<E>().chunk_shape())>>
{
    static const auto& run(const xexpression<E>& e)
    {
        return e.derived_cast().chunk_shape();
    }
};

And then

        template <class E>
        xchunked_array(const xexpression<E>& e)
        {
            xchunked_array<chunk_type> arr(e, get_chunk_shape<E>::run(e));
            *this = arr;
        }