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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T11:16:32+00:00 2026-05-13T11:16:32+00:00

I wrote an array_length function like this: int array_length(int a[]){ return sizeof(a)/sizeof(int); } However

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I wrote an array_length function like this:

int array_length(int a[]){
    return sizeof(a)/sizeof(int);
}

However it is returning 2 when I did

unsigned int len = array_length(arr);
printf ("%i" , len);

where I have

int arr[] = {3,4,5,6,1,7,2};
int * parr = arr;

But when I just do

int k = sizeof(arr)/sizeof(int);
printf("%i", k);

in the main function, it returns 7.

What is the correct way to write the array_length function and how do I use it?

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  1. Editorial Team
    Editorial Team
    2026-05-13T11:16:33+00:00Added an answer on May 13, 2026 at 11:16 am

    Computing array lengths, in C, is problematic at best.

    The issue with your code above, is that when you do:

    int array_length(int a[]){ 
        return sizeof(a)/sizeof(int); 
    } 
    

    You’re really just passing in a pointer as “a”, so sizeof(a) is sizeof(int*). If you’re on a 64bit system, you’ll always get 2 for sizeof(a)/sizeof(int) inside of the function, since the pointer will be 64bits.

    You can (potentially) do this as a macro instead of a function, but that has it’s own issues… (It completely inlines this, so you get the same behavior as your int k =... block, though.)

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