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Home/ Questions/Q 6815181
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T20:46:36+00:00 2026-05-26T20:46:36+00:00

The following code, in global scope, doesn’t compile: const char *one = 1; const

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The following code, in global scope, doesn’t compile:

const char *one = "1";
const char *two = "2";
char *nums[2] = {one, two};

The error message is “initializer element not constant” – which surprises me, since the variables one and two are both declared as constant. Making nums const doesn’t fix the problem. Declaring nums with string literals (char *nums[2] = {"1", "2"};) does fix the problem, but for readability reasons, I’d rather not do it this way in my actual code.

Is there a decent way to get this working?

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  1. Editorial Team
    Editorial Team
    2026-05-26T20:46:36+00:00Added an answer on May 26, 2026 at 8:46 pm

    C does not allow global initialization from variables, even if those are themselves const. By comparison to C++, C has a much stricter notion of a “constant expression”.

    At present, one is a mutable pointer, so it cannot possibly be considered a constant expression, but even the more correct const char * const one = "1"; wouldn’t do in C. (It’d be fine in C++.)

    You’ll have to say:

    const char * nums[2] = { "1", "2" };
    
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