Why do I get an error about the initializer not being a constant?

Question

I am using the following code.

const int X_ORIGIN = 1233086;             
const int Y_ORIGIN = -4728071;              
const int Z_ORIGIN = 4085704;
const int xyzOrigin[NUM_DIMENSIONS] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN};

When I compile it, GCC gives me the following error.

Transformations.h:16:1: error: initializer element is not constant

What does that mean? How can I fix my code?

Why not create a struct which contains the three positions, rather than an array? Seems a bit counter-intuitive... — Nathan White, Jul 18 '12 at 12:17
corollary to what @nathanwhite said, John Carmack said: https://twitter.com/ID_AA_Carmack/status/219884864185446402 — Michael Buen, Jul 18 '12 at 12:52

score 20 · Answer 1 · answered Jul 18 '12 at 13:24

20

Often people are mislead by the naming of the keyword const, implying something of a constant value that can't be changed. In C at least, it means readonly. const qualified objects at file scope are not having the proper constness to serve as array initializers.

As an example for non-constant constness, it is perfectly ok to declare

 const volatile unsigned int milliseconds_since_boot;

being a value that gets updated from outside the compiler's control (think HW register) and that you are not allowed to assign to, i.e. it is readonly.

answered Jul 18 '12 at 13:24

Jens

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Why the downvote? If there's a problem, please give me the opportunity to remedy. – Jens Jul 18 '12 at 15:22
1

Not my DV, but you could elaborate a bit cloaser on OPs actual problem. The example is hard to comprehend for beginners who might not even know what a "HW register" is. – too honest for this site Sep 12 '16 at 11:26

score 14 · Accepted Answer · answered Jul 18 '12 at 12:16

14

You can't do this at global scope in C, only at local scope, i.e. within a function:

#define NUM_DIMENSIONS 3

const int X_ORIGIN = 1233086;             
const int Y_ORIGIN = -4728071;              
const int Z_ORIGIN = 4085704;

const int xyzOrigin[NUM_DIMENSIONS] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN}; // FAIL

void foo(void)
{
    const int xyzOrigin[NUM_DIMENSIONS] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN}; // OK
}

Alternatively you could compile the code as C++ rather than C.

answered Jul 18 '12 at 12:16

Paul R

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That define rather than const int is important. Darn size cannot be any type of variable (even a constant) apparently. – MrMesees Jul 02 '17 at 09:06
Only for automatic storage class which means within a function **and not static (or C11 _Thread_local)**. – dave_thompson_085 Aug 05 '17 at 09:09

score 6 · Answer 3 · answered Jul 18 '12 at 12:17

6

I'm not a proper programmer ;) but I'd do this:

#define X_ORIGIN (1233086)
#define Y_ORIGIN (-4728071)
#define Z_ORIGIN (4085704)
const int xyzOrigin[NUM_DIMENSIONS] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN};

That way it's just a text-substitution. If the compiler still spits the dummy at least you're a step closer to knowing where the issue is.

answered Jul 18 '12 at 12:17

John U

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@nathanwhite Care to elaborate? The macros solve the problem just fine! (Assuming `NUM_DIMENSIONS` is defined somewhere). – Jens Jul 18 '12 at 13:18
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@nathanwhite Those are precisely the causes for the "initializer element is not constant" error(s). Having `NUM_DIMENSIONS` as a variable (const qualified or not), causes a "variably modified »xyzOrigin« at file scope". – Daniel Fischer Jul 18 '12 at 16:48
1

Better still would be to remove `NUM_DIMENSIONS` altogether and let the array size itself to the number of elements provided. – Leushenko Jun 07 '13 at 19:01

score 4 · Answer 4 · answered Jul 18 '12 at 12:27

4

As an alternative, this would also work in this case:

enum { X_ORIGIN = 1233086,
       Y_ORIGIN = -4728071,
       Z_ORIGIN = 4085704 };

const int xyzOrigin[] = { X_ORIGIN, Y_ORIGIN, Z_ORIGIN };

int main()
{
    return 0;
}

answered Jul 18 '12 at 12:27

hmjd

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score 2 · Answer 5 · edited May 23 '17 at 12:17

In C language objects with static storage duration has to be initialized with constant expressions or with aggregate initializers containing constant expressions. --Answer of AndreyT

After reading, You must have the knowledge that NUM_DIMENSIONS, If it has the const-qualification, isn't a constant! Then you can't initializate your array this way.

For use this code:

const int xyzOrigin[NUM_DIMENSIONS] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN};

You should use: #define NUM_DIMENSIONS 3 or you could just declare without any variable inside the square brackets const int xyzOrigin[] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN};

score 0 · Answer 6 · answered Jul 18 '12 at 12:19

As triclosan said:

main()
{
    const int X_ORIGIN = 1233086;
    const int Y_ORIGIN = -4728071;
    const int Z_ORIGIN = 4085704;
    const int xyzOrigin[] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN};
}

this works fine...

or, if you know the dimensions beforehand, this:

#define DIM 3

main()
{
    const int X_ORIGIN = 1233086;
    const int Y_ORIGIN = -4728071;
    const int Z_ORIGIN = 4085704;
    const int xyzOrigin[DIM] = {X_ORIGIN, Y_ORIGIN, Z_ORIGIN};
}

Why do I get an error about the initializer not being a constant?

6 Answers6

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