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Home/ Questions/Q 657365
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T22:48:57+00:00 2026-05-13T22:48:57+00:00

Why does the array a not get initialized by global variable size ? #include<stdio.h>

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Why does the array a not get initialized by global variable size?

#include<stdio.h>

int size = 5;

int main()
{
    int a[size] = {1, 2, 3, 4, 5};
    printf("%d", a[0]);

    return 0;
}

The compilation error is shown as

variable-sized object may not be initialized

According to me, the array should get initialized by size.

And what would be the answer if I insist on using global variable (if it is possible)?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T22:48:58+00:00Added an answer on May 13, 2026 at 10:48 pm

    In C99, 6.7.8/3:

    The type of the entity to be
    initialized shall be an array of
    unknown size or an object type that is
    not a variable length array type.

    6.6/2:

    A constant expression can be evaluated
    during translation rather than runtime

    6.6/6:

    An integer constant expression
    shall have integer type and shall only
    have operands that are integer
    constants, enumeration constants,
    character constants, sizeof
    expressions whose results are integer
    constants, and floating constants that
    are the immediate operands of casts.

    6.7.5.2/4:

    If the size is an integer constant
    expression and the element type has a
    known constant size, the array type is
    not a variable length array type;
    otherwise, the array type is a
    variable length array type.

    a has variable length array type, because size is not an integer constant expression. Thus, it cannot have an initializer list.

    In C90, there are no VLAs, so the code is illegal for that reason.

    In C++ there are also no VLAs, but you could make size a const int. That’s because in C++ you can use const int variables in ICEs. In C you can’t.

    Presumably you didn’t intend a to have variable length, so what you need is:

    #define size 5
    

    If you actually did intend a to have variable length, I suppose you could do something like this:

    int a[size];
    int initlen = size;
    if (initlen > 5) initlen = 5;
    memcpy(a, (int[]){1,2,3,4,5}, initlen*sizeof(int));
    

    Or maybe:

    int a[size];
    for (int i = 0; i < size && i < 5; ++i) {
        a[i] = i+1;
    }
    

    It’s difficult to say, though, what “should” happen here in the case where size != 5. It doesn’t really make sense to specify a fixed-size initial value for a variable-length array.

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