c++ how to initialize const elements of an array

Question

i need a way to initialize const elements of an array for the program i am currently working on. The problem is that i have to initialize these elements with a function, there is no way to do it like this:

const int array[255] = {1, 1278632, 188, ...};

because its alot of data i have to generate. What i tried is to memcpy data to the const int's but that can't work and hasn't worked.

 const int array[255];

 void generateData(){
      for(int i = 0; i < 255; i++) {
          initializeSomehowTo(5, array[i]);
      }
 }

I hope you understand what i am trying, sorry if i doubled the question, i must have overlooked it.

You can't do const int array[255] = initializeMyArray(), but if you used std::array, you could do const std::array<int, 255> = initializeMyArray() — Justin
– Justin, Commented Jun 5, 2017 at 22:43

Arun · Accepted Answer · 2017-06-05 23:01:40Z

20

How about this?

#include <array>
typedef std::array<int, 255> Array;

const Array array = generateData();

Array generateData(){
    Array a;
    for(int i = 0; i < a.size(); i++) {
        initializeSomehowTo(a[i]);
    }
    return a;
}

edited Jun 5, 2017 at 23:01

answered Jun 5, 2017 at 22:45

Arun

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1 Comment

Darius Duesentrieb Over a year ago

this works great, only thing is that const int array[10][10][10]; looks really scary: const std::array<std::array<std::array<int, 10>, 10>, 10> Array; ;-)

Dietmar Kühl · Accepted Answer · 2017-06-05 22:55:26Z

The easiest approach is to get the filled array from a function and use that to initialize your const (or constexpr) object. However, built-in arrays can't be copied but std::array<T, N> be:

std::array<T, 255> array = initializeData();

If you need a built-in array, I can imagine initializing a static member of a class (template, actually) where the index is expanded from indices expanded from an std::make_index_sequence<255> and used as positional argument in the array, i.e., something along these lines:

#include <algorithm>
#include <iostream>
#include <iterator>
#include <utility>

int some_function(std::size_t i) { return i; }

template <typename> struct initialized_array_base;
template <std::size_t... I>
struct initialized_array_base<std::index_sequence<I...>> {
    static const int array[sizeof...(I)];
};

template <std::size_t... I>
int const initialized_array_base<std::index_sequence<I...>>::array[sizeof...(I)]
    = { some_function(I)... };

struct initialized_array
    :initialized_array_base<std::make_index_sequence<256>> {
};

int main() {
    std::copy(std::begin(initialized_array::array),
              std::end(initialized_array::array),
              std::ostream_iterator<int>(std::cout, " "));
    std::cout << '\n';
}

Amadeus · Accepted Answer · 2017-06-05 23:00:11Z

3

You can create a writable array, initialize it, and, then, create a const reference to it.

int arry[255];

void generateData(){
  for(int i = 0; i < 255; i++) {
      initializeSomehowTo(5, arry[i]);
  }
}

const int (&array)[255] = arry;

answered Jun 5, 2017 at 23:00

Amadeus

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3 Comments

Arun Over a year ago

This does not stop arry from getting modified, right?

Brady Dean Over a year ago

I think you can also std::static_cast to a const reference

Peter Cordes Over a year ago

An extra level of indirection (for the reference variable that also has static storage) may not optimize away. I would definitely recommend not using this, since returning a std::array<> from an initializer function avoids any performance downside.

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