Strings in java are called by value not reference

Why?

Because String is immutable.

Assert in java

public class Test2 { public static int x; public static int foo(int y) { return y * 2; } public static void main(String [] args) { int z = 5; assert z > 0; /* Line 11 */ assert z > 2: foo(z); /* Line 12 */ if ( z < 7 ) assert z > 4; /* Line 14 */ switch (z) { case 4: System.out.println("4 "); case 5: System.out.println("5 "); default: assert z < 10; } if ( z < 10 ) assert z > 4: z++; /* Line 22 */ System.out.println(z); } } which line is an example of an inappropriate use of assertions?

A. Line 11 B. Line 12 C. Line 14 D. Line 22 Answer: Option D Explanation: Assert statements should not cause side effects. Line 22 changes the value of z if the assert statement is false. Option A is fine; a second expression in an assert statement is not required. Option B is fine because it is perfectly acceptable to call a method with the second expression of an assert statement. Option C is fine because it is proper to call an assert statement conditionally.

Reference :http://www.indiabix.com/online-test/java-programming-test/61

Output of program

public class Test { public static int y; public static void foo(int x) { System.out.print("foo "); y = x; } public static int bar(int z) { System.out.print("bar "); return y = z; } public static void main(String [] args ) { int t = 0; assert t > 0 : bar(7); assert t > 1 : foo(8); /* Line 18 */ System.out.println("done "); } }

A. bar B. bar done C. foo done D. Compilation fails Answer: Option D Explanation: The foo() method returns void. It is a perfectly acceptable method, but because it returns void it cannot be used in an assert statement, so line 18 will not compile. Learn more problems on : Assertions Discuss about this problem : Discuss in Forum http://www.indiabix.com/online-test/java-programming-test/61

Assert in java

Refer to

http://www.javatpoint.com/assertion-in-java

Thread And Runnable both have run method which one will be called?

class MyThread extends Thread { MyThread() {} MyThread(Runnable r) {super(r); } public void run() { System.out.print("Inside Thread "); } } class MyRunnable implements Runnable { public void run() { System.out.print(" Inside Runnable"); } } class Test { public static void main(String[] args) { new MyThread().start(); new MyThread(new MyRunnable()).start(); } }

A. Prints "Inside Thread Inside Thread" B. Prints "Inside Thread Inside Runnable" C. Does not compile D. Throws exception at runtime Answer: Option A Explanation: If a Runnable object is passed to the Thread constructor, then the run method of theThread class will invoke the run method of the Runnable object. In this case, however, the run method in the Thread class is overridden by the run method in MyThread class. Therefore the run() method in MyRunnable is never invoked. Both times, the run() method in MyThread is invoked instead. Reference: http://www.indiabix.com/online-test/java-programming-test/61

IllegalThreadStateException

class MyThread extends Thread { public static void main(String [] args) { MyThread t = new MyThread(); t.start(); System.out.print("one. "); t.start(); System.out.print("two. "); } public void run() { System.out.print("Thread "); } }

A. Compilation fails B. An exception occurs at runtime. C. It prints "Thread one. Thread two." D. The output cannot be determined. Answer: Option B Explanation: When the start() method is attempted a second time on a single Thread object, the method will throw an IllegalThreadStateException (you will not need to know this exception name for the exam). Even if the thread has finished running, it is still illegal to call start() again.

short circuit operator in java

class BoolArray 
{
    boolean [] b = new boolean[3];
    int count = 0;

    void set(boolean [] x, int i) 
    {
        x[i] = true;
        ++count;
    }

    public static void main(String [] args) 
    {
        BoolArray ba = new BoolArray();
        ba.set(ba.b, 0);
        ba.set(ba.b, 2);
        ba.test();
    }

    void test() 
    {
        if ( b[0] && b[1] | b[2] )
            count++;
        if ( b[1] && b[(++count - 2)] )
            count += 7;
        System.out.println("count = " + count);
    }
}

Result count = 3

Why?

The reference variables b and x both refer to the same boolean array. count is incremented for each call to the set() method, and once again when the first if test istrue. Because of the && short circuit operator, count is not incremented during the second if test.

Output of program

class SC2 
{
    public static void main(String [] args) 
    {
        SC2 s = new SC2();
        s.start();
    }

    void start() 
    {
        int a = 3;
        int b = 4;
        System.out.print(" " + 7 + 2 + " ");
        System.out.print(a + b);
        System.out.print(" " + a + b + " ");
        System.out.print(foo() + a + b + " ");
        System.out.println(a + b + foo());
    }

    String foo() 
    {
        return "foo";
    }
}

Result

72 7 34 foo34 7foo

Description:

Because all of these expressions use the + operator, there is no precedence to worry about and all of the expressions will be evaluated from left to right. If either operand being evaluated is a String, the + operator will concatenate the two operands; if both operands are numeric, the + operator will add the two operands.