Is it possible to set the value of elements in a constexpr array after declaration?

Question

Is it possible to declare a const array (possibly constexpr) at one point, then define it at another place, one element at a time?

E.g.

extern constexpr int myArray[100];


myArray[0] = myConstexprFunction(0);
myArray[1] = myConstexprFunction(1);
// ...
myArray[100] = myConstexprFunction(100);

What I'm trying to do will need something like this. Maybe it's possible using things like: http://b.atch.se/posts/constexpr-counter/

But if this technique is going to be illegal in the next C++ standard (I hope not) I would like to use a safer one.

[EDIT] how about relaxing some requirements.. let's say that I want to do something like this:

constexpr int myConstExprFunction(int arg) { return arg + 100;}
// other code...
constexpr int a = myConstExprFunctionBegin(10);
constexpr int b = myConstExprFunction(20);
constexpr int c = myConstExprFunction(30);
constexpr int d = myConstExprFunctionEnd(40);

what I would like to have is that the myConstExprFunctionEnd is able to generate a final array with the values created by the previous functions. Everything at compile time of course.

[EDIT2] C++11 solutions very welcomed

@RichardCritten yes, sure I've tried, but I get different kind of errors ( .. has been previously declared.. / conflicting declaration ... etc.). The only way I could do something like that has been in a recursive way using template metaprogramming — user3770392, Apr 04 '17 at 11:16

score 6 · Answer 1 · edited May 23 '17 at 12:32

6

The requirement of constexpr of the recent C++ is very relaxed, so you could just write:

// requires C++17:
constexpr auto myArray = [] {
    std::array<int, 100> result {};
    for (size_t i = 0; i < 100; ++ i) {
        result[i] = i * i;
    }
    return result;
}();

Note I used std::array<int, 100> instead of int[100] because a function cannot return a C array.

The above code requires C++17 for two reasons:

constexpr lambda
The mutable operator[] is not constexpr before C++17

Issue 1 can be easily worked-around using a separate constexpr function. Issue 2 can only be solved by defining your own array wrapper.

// requires C++14:

template <typename T, size_t n>
struct ConstexprArray {
    T data[n];
    constexpr ConstexprArray() : data{} {}
    constexpr T& operator[](size_t i) { return data[i]; }
};

constexpr auto initialize_my_array() -> ConstexprArray<int, 100> {
    ConstexprArray<int, 100> result {};
    for (size_t i = 0; i < 100; ++ i) {
        result[i] = i * i;
    }
    return result;
}

constexpr auto myArray = initialize_my_array();

edited May 23 '17 at 12:32

Community

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answered Apr 04 '17 at 12:01

kennytm

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still not exactly what I was looking for but very interesting! have you read the second part of the post (the edit) ? – user3770392 Apr 04 '17 at 12:10
@user3770392 How is the second part *not* capable with C++17? – kennytm Apr 04 '17 at 13:12
I was meaning the second part of my post. By the way, I can use just C++11. C++14 elements like indices etc ok, I can simulate them – user3770392 Apr 04 '17 at 13:32
@user3770392 OK I see (yes I do mean the second part of your post). – kennytm Apr 04 '17 at 14:26
ahahaa ok, let's say I meant just C++11 then :) – user3770392 Apr 04 '17 at 14:52

Sumeet · Answer 2 · 2017-04-04T11:49:22.967

4

constexpr declares that it is possible to evaluate the value of the function or variable at compile time.

So the only way you could use it with array is like:

constexpr int myArray[100]{1 , 2 , 3 ,.........};

statements like

myArray[0] = myConstexprFunction(0);

can only be evaluated during runtime. So its not possible.

edited Apr 04 '17 at 11:49

answered Apr 04 '17 at 11:14

Sumeet

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I'm able to fill that array using template metaprogramming techniques, but it is equivalent to creating a { ... } – user3770392 Apr 04 '17 at 11:18

Guillaume Racicot · Accepted Answer · 2021-03-05T22:13:15.177

4

Looking at your edit, I'd just answer no, because the compiler cannot transform a group of variables into an array. It just don't work that way. There isn't any construct in C++ that can take a bunch of declaration, delete them and replace it with another declaration. A source code preprocessor or generator might be able to permit the syntax you seek.

If you're interested in a solution that doesn't require external tooling, you can create a constexpr function that returns an array:

constexpr auto makeMyArray() {
    std::array<int, 100> myArray{};

    myArray[0] = myConstExprFunction(10);
    myArray[1] = myConstExprFunction(20);
    // ...

    return myArray;
}

Then, initialize your array:

constexpr auto myArray = makeMyArray();

edited Mar 05 '21 at 22:13

answered Apr 04 '17 at 14:38

Guillaume Racicot

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interesting, this seems C++17 right? no chance to do something like this outside a constexpr function in C++11? – user3770392 Apr 04 '17 at 14:53
2

Not in C++11. You can however, replace loops or multiple statements with recursive function call. It requires C++14. You cannot have multiple statements in a constexpr function in C++11. – Guillaume Racicot Apr 04 '17 at 14:58
1

I think I can use the comma operator too :) – user3770392 Apr 04 '17 at 15:00
1

Oh maybe! Didn't thought about that! But I doubt you can mutate values with this. – Guillaume Racicot Apr 04 '17 at 15:02
mmmm yes, it's not easy, I have to find a solution though :) – user3770392 Apr 04 '17 at 15:04
1

Try `-std=c++14` – Guillaume Racicot Apr 04 '17 at 15:04
1

eheheh I know how to use that directive, but I can't use it right now... – user3770392 Apr 04 '17 at 15:06
1

In that case, I'd accept a less appealing syntax (put function calls in the array initializer) and leave a comment about how should that be replaced in C++14 – Guillaume Racicot Apr 04 '17 at 15:07
yes I was thinking to do something like this.. problem is however that what I wanted to do probably is impossible in C++11 – user3770392 Apr 04 '17 at 15:14
Note that (at least for MSVC) the array declaration needed {}'s. e.g. `std::array<...> arr{};` – Matt Eding Mar 05 '21 at 21:14
@MattEding I don't think constexpr allow for uninitialized data still. I edited the answer – Guillaume Racicot Mar 05 '21 at 22:13

score 3 · Answer 4 · answered Apr 04 '17 at 11:39

If you want to declare constexpr an array and initialize it's value using a constexpr function... the best I can think is wrap the array in a struct/array and initialize it via a delegate constructor.

The following is a full working C++14 example

#include <utility>
#include <iostream>

constexpr int myConstexprFunction (int i)
 { return i << 1; } // return 2*i

template <std::size_t S>
struct wrapArray
 {
   int const myWrappedArray[S];

   template <int ... Is>
   constexpr wrapArray (std::integer_sequence<int, Is...> const &)
      : myWrappedArray { myConstexprFunction(Is)... }
    { }

   constexpr wrapArray ()
      : wrapArray(std::make_integer_sequence<int, S>())
    { }
 };


int main ()
 {
   constexpr wrapArray<100>  wa100;

   for ( auto i : wa100.myWrappedArray )
      std::cout << i << ", ";

   std::cout << std::endl;
 }

If you need a C++11 code, you have to implement a substitute for std::integer_sequence and for std::make_integer_sequence(). It's not difficult.

yes this is similar to what I did, but is not enough for my purposes. I'll edit my question with some more detail — user3770392, Apr 04 '17 at 11:44

score 1 · Answer 5 · answered Apr 04 '17 at 11:18

1

No.

constexpr variables must be "immediately initialised".

answered Apr 04 '17 at 11:18

Lightness Races in Orbit

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Is it possible to set the value of elements in a constexpr array after declaration?

5 Answers5