Thread: sizeof(arrays)

My main goal is to find out the number of elements in an array. In my program I have something like this:

Code:

int arr1[] = {3,4,6,7};

printf("The array has %d elements.\n", sizeof(arr1)/sizeof(int));

Which works fine. The problem is when I pass the array to a function of mine and use the same print statement. the result is always 1;

the function looks something like this:

Code:

int nth(int[], int[]);


int nth(int arr1[], int arr2[]{
    printf("The array has %d elements.\n", sizeof(arr1)/sizeof(int));

    return arr1[0];
}

What am I missing here? Thanks...

You can't pass arrays as parameters in C. What is actually happening is that you are passing a pointer to the first element of the array (the brackets are just syntactic sugar). You will need to pass the size as another parameter.

When you pass an array to a function, it automatically decays to a pointer type (IE: int*). So, in your case, I assume sizeof(int*) equals sizeof(int), and you therefore get the result of 1.

The closest you could get to what you want is a macro, which also lets you abstract the type of the array:

Code:

#define NTH(A)   (sizeof(A) / sizeof((A)[0]))

Originally Posted by rt454

You can't pass arrays as parameters in C. What is actually happening is that you are passing a pointer to the first element of the array (the brackets are just syntactic sugar). You will need to pass the size as another parameter.

Ya I considered passing the size but I was hoping I could get around it to keep the parameter list down. Oh well thanks for the info.

Ronix, I read about macros like that before but I can't tell exactly what that macro you wrote means. Would you mind explaining it in a little more detail?

Originally Posted by zacs7

> So, in your case, I assume sizeof(int*) equals sizeof(int)
Very bad assumption to make in any case, esp with 64-bit being so common. But anywho.

Yeah. However, Nextstopearth said that his code yielded a value of 1, which means that sizeof(int*) and sizeof(int) are equal.