Convert an int to a char * pointer

itoa.c

/*
 * itoa - convert an int to a str with the given base.
 * My own implementation with LOTS of explanation
 *
 */
char *itoa(int val, int base) {

	// make sure base is supported
	if (base < 2 || base > 16) {
		return NULL;
	}

	// NUMBERS to use for lookups
	static char NUMBERS [] = "0123456789abcdef";

	// use newstr to store newly built string
	static char newstr[32] = { '\0' };

	// starting index
	// leave last char as 0, '\0', etc
	int i = 30;

	// while we still have more of val to process,
	// and while we haven't got the the beginning of
	// our new str (ran out of space)
	while (val && i) {

		// val % base --> remainder is the value of the next digit
		// for example, if val is 14 and we're doing decimal (base 10)
		// then the next_digit would be 4.
		int next_digit = val % base;
		newstr[i] = NUMBERS[next_digit];

		// move to next index in newstr
		// (actually previous index, or next index,
		// but we're going backwars
		i--;

		// move to the next digit in val
		// ex: move from tens digit to ones digit
		val = val / base;
	}

	// return pointer to first letter of newstr
	// since we decremented i (i--) at the end of
	// the loop, its now pointing at nothing.
	// we have to return address of newstr[i + 1] to
	// get the last char we assigned it (which should be
	// what the newstr starts with, if that makes sense)
	return &newstr[i + 1];
}