Array length counting anomaly

Question

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

#include 
int l         


        

Answer 1

There is not going to be a zero at the end of your array unless you put one there! A literal char array defined using a character string does, indeed, have such a sentinel value, but they're special; other arrays have no equivalent.

The len() function you're trying to write cannot be written for general arrays in C -- there's no way to determine the size of a dynamic array without using the (undocumented, platform-specific) internals of the memory allocator. If it was important to do this for your application, you could only do it if it were possible for you to add a zero at the end of every array yourself, explicitly.

Answer 2

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.

Answer 3

Unlike strings, normal arrays do not terminate with a null byte 0x00. The reason strings use this is because arrays have no concept of length; arrays are merely contiguous pieces of memory, it is up to you to keep track of the length of arrays.

Answer 4

You cannot count arrays like that. Only strings are null terminated. If you want that to work reliably, you will need to add an additional element to your array that contains '\0'. But be sure to remember to take into account that your length will no be one larger then the true length because of that '\0'