variadic template for forward list initialization of std::array

Question

I have a geometry structure describing a point in N dimensions

template <typename T, std::size_t N>
class point : public std::array<T, N>

Among the many constructors / methods I'd like to add to this class, I cannot figure out how to build a variadic constructor. I've tried

template<typename ...Args>
point(Args&&...t) : std::array<T,N>(std::forward<Args>(t)...)
{
}

but that doesn't work

std::array<int,5> p1 = {0, 1, 1, 2, 3}; // OK
point<int,5>      p2 = {0, 1, 1, 2, 3}; // ERROR

how can I define such a constructor ?

Perhaps you're looking for a constructor taking std::initializer_list? — John Zwinck
– John Zwinck, Commented Mar 4, 2015 at 1:58
What is your question? The last snippet should compile, combine that with std::forward and you have your original desired constructor. — Praetorian
– Praetorian, Commented Mar 4, 2015 at 2:24
@Amxx Then you should post it as an answer. The reason you need to use braces instead of parentheses is because std::array is an aggregate and has no user defined constructors, so you must perform aggregate initialization. — Praetorian
– Praetorian, Commented Mar 4, 2015 at 2:36

Amxx · Accepted Answer · 2015-03-04 02:42:21Z

3

This behaviour comes from the specificity of std::array. Being an agregate without common constructor means you can't initialize it using std::forward or std::initializer_list

On should rather use double braces like so:

template<typename ...Args>
point(Args&&... args) : std::array<T,N>{{args...}}

answered Mar 4, 2015 at 2:42

Amxx

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