initialize an array of compile-time constant size with incrementing numbers

Question

I have an array whose size is set using a compile-time constant (a pre-processor #define in my case). I need to initialize it using consecutive numbers at compile-time. How can I do this?

Simplified example:

#define ARR_SZ 5
struct C {
  C(int a) : a(a) {}
  int a;
};
C arr[ARR_SZ] = {{0},{1},{2},{3},{4}}; // This needs to adapt to any number

I can use C++11, but not newer (although I would be interested to learn of newer techniques even if I can't use them for this project)

[`std::iota`](https://en.cppreference.com/w/cpp/algorithm/iota)? — Some programmer dude, Jan 21 '19 at 09:16
`std::make_index_sequence` (C++14) (can be implemented in C++11). — Jarod42, Jan 21 '19 at 09:18
@StoryTeller In the actual code it is an `std::array`, as well as being of a class type with no default constructor and only a constructor that accepts an integer. I was trying to simplify — Baruch, Jan 21 '19 at 09:19
The use of a class without a default constructor could make a big difference. Always try to create a [mcve] to show us, don't simplify to much. — Some programmer dude, Jan 21 '19 at 09:20
The answers in the "duplicate" question are about assigning to an array after it was created. I need to initialize it at deceleration — Baruch, Jan 21 '19 at 09:25
@P.W No, but out of curiosity, what does Boost have to offer here? — Baruch, Jan 21 '19 at 10:02
@Baruch: If the number of elements are decided at compile time then not much difference between vector and array. — P.W, Jan 21 '19 at 11:16
@P.W No, this is an embedded system with no dynamic allocation allowed or even supported — Baruch, Jan 21 '19 at 12:15

score 3 · Accepted Answer · answered Jan 21 '19 at 09:46

3

C++14 code (because of std::integer_sequence):

#include <type_traits>
#include <array>

#define ARR_SZ 5
struct C {
  C(int a) : a(a) {}
  int a;
};

template<int ...Is>
auto make_C_arr(std::integer_sequence<int, Is...>) -> std::array<C, sizeof...(Is)> {
    return {{ {Is}... }};
}

auto arr = make_C_arr(std::make_integer_sequence<int, ARR_SZ>{});

int main () {

}

std::integer_sequence and the like are implementable in C++11 however as noted in a comment, so substituting the standard version for a home-brewed one will give a C++11 specific solution.

answered Jan 21 '19 at 09:46

StoryTeller - Unslander Monica

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Can this be done without the extra function call (`make_C_arr`)? It just looks ugly – Baruch Jan 21 '19 at 13:20
@Baruch - No. Not this. Something needs to unpack. You can try the preprocessor instead if you prefer. – StoryTeller - Unslander Monica Jan 21 '19 at 13:22
Why is the double-bracket needed? (`{{ {Is}.. }}` instead of `{ {Is}.. }`) – Baruch Jan 21 '19 at 14:01
1

@Baruch - https://stackoverflow.com/questions/11400090/why-is-the-c-initializer-list-behavior-for-stdvector-and-stdarray-differen – StoryTeller - Unslander Monica Jan 21 '19 at 14:04

score 1 · Answer 2 · answered Jan 21 '19 at 12:12

Since boost was mentioned in the comment section, here's another totally different solution based on Boost.PP. It's also entirely C++03.

#include <boost/preprocessor/repetition/repeat.hpp>
#include <boost/preprocessor/punctuation/comma_if.hpp>

#define ARR_SZ 5
struct C {
  C(int a) : a(a) {}
  int a;
};

#define INIT(z, n, d) BOOST_PP_COMMA_IF(n) C(n)

C arr[ARR_SZ] = { BOOST_PP_REPEAT(ARR_SZ, INIT, ?) };


int main () {

}

BOOST_PP_REPEAT will expand to INIT(z, 0, ?) ... INIT(z, 4, ?). The z is not relevant to our goal, and the ? token is just a placeholder. Since INIT in turn expands to C(n) for n from 0 to 4 (comma delimited), we get an initializer for a regular C style array.

initialize an array of compile-time constant size with incrementing numbers

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