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:

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

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

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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:

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


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

    return arr1[0];
}

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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:

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

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Quote:

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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.

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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?

> 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.

> 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?
Even though I'm not Ronix... The macro aspect is not important to portray the concept. Basically it tells you how many elements in a one-dimension array. Since sizeof() returns the amount of "bytes" a specific type/object takes up then you need to divide the total bytes occupied by the array by the size of each array element.

Ie given an array that occupies 16 bytes (sizeof array) and given each element occupies 4 bytes (sizeof *array or sizeof array[0]) then there are,
= 16 / 4
or
= sizeof(array) / sizeof(array[0]);
Elements in the array.

Quote:

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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.

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