Get length of string array of unknown length

Question

I have this function:

int setIncludes(char *includes[]);

I don't know how many values includes will take. It may take includes[5], it may take includes[500]. So what function could I use to get the length of includes?

Answer 1

There is none. That's because arrays will decay to a pointer to the first element when passing to a function.

You have to either pass the length yourself or use something in the array itself to indicate the size.

---

First, the "pass the length" option. Call your function with something like:

int setIncludes (char *includes[], size_t count) {
    // Length is count.
}
:
char *arr[] = {"Hello,", "my", "name", "is", "Pax."};
setIncludes (arr, sizeof (arr) / sizeof (*arr));
setIncludes (arr, 2); // if you don't want to process them all.

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A sentinel method uses a special value at the end to indicate no more elements (similar to the \0 at the end of a C char array to indicate a string) and would be something like this:

int setIncludes (char *includes[]) {
    size_t count = 0;
    while (includes[count] != NULL) count++;
    // Length is count.
}
:
char *arr[] = {"Hello,", "my", "name", "is", "Pax.", NULL};
setIncludes (arr);

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Another method I've seen used (mostly for integral arrays) is to use the first item as a length (similar to Rexx stem variables):

int setIncludes (int includes[]) {
    // Length is includes[0].
    // Only process includes[1] thru includes[includes[0]-1].
}
:
int arr[] = {4,11,22,33,44};
setIncludes (arr);

Answer 2

You have two options:

1. You can include a second parameter, similar to:

   int main(int argc, char**argv)

2. ... or you can double-null terminate the list:

   char* items[] = { "one", "two", "three", NULL }

Answer 3

There is no way to simply determine the size of an arbitrary array like this in C. It requires runtime information that is not provided in a standard way.

The best way to support this is to take in the length of the array in the function as another parameter.