class with initialization list initiated std::array member variable

Question

In c++17 you can do

std::array a{ "one", "two", "three" };

and get a std::array<const char*, 3>. Awesome!

Because array member variables must have their size specified in the class declaration this is not usable as a class variable. I would like to create a class with one of these arbitrarily long initialization list initialized arrays and one or more constexpr methods. I would expect the only constructor to take the initialization list as an argument. Is there a way to do this?

std::array is usable as class field just fine. But if you need variable size array then you need std::vector — user7860670
– user7860670, Commented Oct 27, 2018 at 20:56
std::vercot is unsuitable due to compile-time initialization requirements. The array size would be fixed per instance, but may be different in different instances. — Michael Surette
– Michael Surette, Commented Oct 27, 2018 at 21:00
@MichaelSurette Oh, it's suitable alright. It has stood the test of time just fine. — Ron
– Ron, Commented Oct 27, 2018 at 21:01
If you know array size at compile time then what is the problem with using std::array? — user7860670
– user7860670, Commented Oct 27, 2018 at 21:01
I would like to have the size of the initialization list be the size of the array, removing the requirement to count the entries, which may be a source of errors, especially on updates. — Michael Surette
– Michael Surette, Commented Oct 27, 2018 at 21:04

Pezo · Accepted Answer · 2018-10-27 21:22:55Z

6

You can use a class templated on the size of the array (and the element type if you want), that has a variadic constructor. Something like this:

template <typename T, std::size_t N>
struct custom_array {
    std::array<T, N> data;

    template <typename... Args>
    constexpr explicit custom_array(Args... args) : data{args...} {
    }
};

To be able to deduce N you also need a deduction guide like this:

template <typename... Args>
explicit custom_array(Args... args) -> custom_array<std::common_type_t<Args...>, sizeof...(Args)>;

You can find a complete example on Compiler Explorer.

answered Oct 27, 2018 at 21:22

Pezo

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4 Comments

Michael Surette Over a year ago

Thank you for this worthy effort, it does however still suffer from the same issue as std::array in that you have to declare the initialization list in the class definition. This necessitates a separate class for each instance. It is an improvement in that the methods can be integrated into the template. Definitely worth studying to see if I can get around this limitation.

Pezo Over a year ago

So you want std::arrays of different sizes in the same non-template class? I'm afraid that's not possible.

Michael Surette Over a year ago

I don't mind a template class, just not a whole new class for each initialization list. If what I'm looking for is not possible, then I am a bit disappointed, but it is what it is.

Pezo Over a year ago

It's a new instantiation for each type-size combination, that's true. It's not possible because arrays of different sizes are entirely different types. What you could do is to put most of your logic in a non-template base class and use virtual member functions to abstract the size-dependent parts of the concrete classes away.

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