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Threads - Programming

Q1:

What will be the output of the program? 
public class Test 
{
    public static void main (String [] args) 
    {
        final Foo f = new Foo();
        Thread t = new Thread(new Runnable() 
        {
            public void run() 
            {
                f.doStuff();
            }
        });
        Thread g = new Thread() 
        {
            public void run() 
            {
                f.doStuff();
            }
        };
        t.start();
        g.start();
    }
}
class Foo 
{
    int x = 5;
    public void doStuff() 
    {
        if (x < 10) 
        {
            // nothing to do
            try 
            {
                wait();
                } catch(InterruptedException ex) { }
        } 
        else 
        {
            System.out.println('x is ' + x++);
            if (x >= 10) 
            {
                notify();
            }
        }
    }
}

A The code will not compile because of an error on notify(); of class Foo.

B The code will not compile because of some other error in class Test.

C An exception occurs at runtime.

D It prints "x is 5 x is 6".

ANS:A - The code will not compile because of an error on notify(); of class Foo.

C is correct because the thread does not own the lock of the object it invokes wait() on. If the method were synchronized, the code would run without exception. A, B are incorrect because the code compiles without errors. D is incorrect because the exception is thrown before there is any output.