Question

The count is returning unpredictable results. Sometimes they are right. Sometimes totally weird. Anyone can tell me what is wrong?

#include <stdio.h>
int len(int[]);

int main (int argc, const char * argv[])
{
    int a[]={1,2,3,4,5,6,7,8};
    int* i = a;
    printf("length is %d",(len(i)));
    return 0;
}

int len(int* a){
    int count = 0;
    for (; *a!='\0'; a++) {
        count++;
    }
    return count; 
}

Answer 1

I think you’re confused between C strings (arrays of char) and other arrays. It’s a convention that C strings are terminated with a null character ('\0'), but not all arrays (even char arrays) are terminated this way.

The general convention is to either store the length of an array somewhere, or to use a sentinel value at the end of the array. This value should be one that won’t come up inside the array – eg '\0' in strings, or -1 in an array of positive ints.

Also, if you know that a is an int array (and not a pointer to an int array), then you can use:

size_t length = sizeof(a) / sizeof(a[0]);

So you could do:

int a[] = {1,2,3,4,5,6,7,8};
size_t length = sizeof(a) / sizeof(a[0]); 

// In this case, sizeof(a[0]) 
// is the same as sizeof(int), because it's an int array.

But you can’t do:

int *a = malloc(sizeof(int) * 10);
size_t length = sizeof(a) / sizeof(a[0]); // WRONG! 

That last example will compile, but the answer will be wrong, because you’re getting the size of a pointer to the array rather than the size of the array.

Note that you also can’t use this sizeof to read the size of an array that’s been passed into a function. It doesn’t matter whether you declare your function len(int *a) or len(int a[]) – a will be a pointer, because the compiler converts arrays in function arguments to be a pointer to their first element.