Bitops.java

public class Bitops {
	
	// Takes a character and returns the number of ones
	// contained in its binary representation
	static int numOnes(char c) { 
		int hex = (int) c;
		int sum = 0;
		// we want unsigned comparison
		// this works fine if java uses 2s complement (it does I think).
		while (hex > 0) { 
			sum += hex & 0x1; // loops a maximum of 8 times
			hex = hex >> 1; // signed shift.
		}
		return sum;
	}

	// n = length of the string str
	// O(n * 8) where 8 is the number of bits per character.
	static int numOnesInString(String str){
		int sum = 0;
		while (str.length() > 0){
			char firstChar = str.charAt(0);
			sum += numOnes(firstChar);
			str = str.substring(1);
		}
		return sum;
	}
	
	public static void main(String[] args) {
		System.out.println(numOnesInString("  ") == 2); // Each space is 0x20, which has one 1 bit in it.
		System.out.println(numOnesInString("\0") == 0); // Null character is 0x00, which has no 1 bits in it.
		System.out.println(numOnesInString("H") == 2); // H is 01001000, which has two 1 bits in it.
		System.out.println(numOnesInString("z") == 5); // z is 01111010, which has five 1 bits in it.
		System.out.println(numOnesInString("Hi") == 6);	// i is 01101001, which has four 1 bits in it (then add two for H).

		String str = "\n# of ones in \0this string is what?";
		System.out.println(str);
		System.out.println(numOnesInString(str));
	}
}