How to typedef a std::array with a unspecified size?

Question

I want to write some variables like

std::array<double, array_num> a;

where array_num is a const int representing the length of the array. But it's long and I want to create an alias for it:

typedef std::array<double, array_num> my_array;

Is it right? How can I use my_array like my_array<3>?

Answer 1

What you need is an alias template:

template <size_t S>
using my_array = std::array<double, S>;

You cannot directly make a typedef template, see this post.

^{size_t is the type of the second template parameter std::array takes, not int.}

^{Now that you know about using, you should be using that. It can do everything what typedef does, plus this. Also, you read it from left to right with a nice = sign as a delimiter, as opposed to typedef, which might hurt your eyes sometimes.}

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Let me add two more examples of use:

template <typename T>
using dozen = std::array<T, 12>;

And if you wanted to create an alias for std::array, such as it is, you'd need to mimic its template signature:

template <typename T, size_t S>
using my_array = std::array<T, S>;

- because this is not allowed:

using my_array = std::array;