import java.security.*;
import java.nio.ByteBuffer;
import java.util.Random;

public class SHATest {
    protected static int toSHA1(byte[] bytes) {
	MessageDigest md = null;
	try {
	    md = MessageDigest.getInstance("SHA-1");
	}
	catch(NoSuchAlgorithmException e) {
	    e.printStackTrace();
	    System.exit(1);
	}
	ByteBuffer bb = ByteBuffer.wrap(md.digest(bytes));
	return bb.getInt();
    }

    protected static String intAsBinaryString(int num) {
	String s = "";
	for (int i = 0; i < 4; i++) {
	    // this gets the ith byte
	    byte ithByte = (byte) ((num >> 8 * i) & 0xFF);
	    s = byteAsBinaryString(ithByte) + s;
	}
	return s;
    }

    protected static String byteAsBinaryString(byte num) {
	String s = "";
	for (int i = 0; i < 8; i++) {
	    byte mask = (byte) (1 << i);
	    if ((num & mask) != 0) {
		s = "1" + s;
	    } else {
		s = "0" + s;
	    }
	}
	return s;
    }
    
    protected static <T> byte[] keyToBytes(T key) {
	int hc = key.hashCode();
	byte[] bytes = new byte[4];
	/* Fill in this part.
         * The goal is to turn the int hc into a byte array.
         * You will need to extract each of the four bytes
         * and put them into the byte[] declared above.
         * Looking around at other code in this program should
         * give you some very helpful hints.
         */
	return bytes;
    }

    protected static String randomString() {
	Random r = new Random();
	String s = "";
	for (int i = 0; i < 8; i++) {
	    // 97 is ASCII 'a', so 97 + k is a letter from a-z.
	    int k = r.nextInt(26) + 97;
	    char c = (char)k;
	    s += c;
	}
	return s;
    }

    public static void main(String[] args) {
	String key = randomString();
	System.out.println("key: " + key);
	System.out.println("hash: " + toSHA1(keyToBytes(key)));
	// is the hash deterministic? yes, if the line above and line below agree.
	System.out.println("hash: " + toSHA1(keyToBytes(key)));
    }
}