Q: What will be the output of the program? int i = 1, j = 10; do { if(i++ > --j) /* Line 4 */ { continue; } } while (i < 5); System.out.println("i = " + i + "and j = " + j); /* Line 9 */

This question is not testing your knowledge of the continue statement. It is testing your knowledge of the order of evaluation of operands. Basically the prefix and postfix unary operators have a higher order of evaluation than the relational operators. So on line 4 the variable i is incremented and the variable j is decremented before the greater than comparison is made. As the loop executes the comparison on line 4 will be: if(i > j) if(2 > 9) if(3 > 8) if(4 > 7) if(5 > 6) at this point i is not less than 5, therefore the loop terminates and line 9 outputs the values of i and j as 5 and 6 respectively. The continue statement never gets to execute because i never reaches a value that is greater than j.

Question 1

What will be the output of the program? int i = 1, j = 10; 
do 
{
    if(i++ > --j) /* Line 4 */
    {
        continue; 
    } 
} while (i < 5); 
System.out.println("i = " + i + "and j = " + j); /* Line 9 */

Accepted Answer

This question is not testing your knowledge of the continue statement. It is testing your knowledge of the order of evaluation of operands. Basically the prefix and postfix unary operators have a higher order of evaluation than the relational operators. So on line 4 the variable i is incremented and the variable j is decremented before the greater than comparison is made. As the loop executes the comparison on line 4 will be: if(i > j) if(2 > 9) if(3 > 8) if(4 > 7) if(5 > 6) at this point i is not less than 5, therefore the loop terminates and line 9 outputs the values of i and j as 5 and 6 respectively. The continue statement never gets to execute because i never reaches a value that is greater than j.

Question 2

What will be the output of the program? String d = "bookkeeper";
d.substring(1,7);
d = "w" + d;
d.append("woo");  /* Line 4 */
System.out.println(d);

Accepted Answer

In line 4 the code calls a StringBuffer method, append() on a String object.

Question 3

What will be the output of the program? System.out.println(Math.sqrt(-4D));

Accepted Answer

It is not possible in regular mathematics to get a value for the square-root of a negative number therefore a NaN will be returned because the code is valid.

Question 4

What will be the output of the program? public class NFE 
{
    public static void main(String [] args) 
    {
    String s = "42";
        try 
        {
            s = s.concat(".5");  /* Line 8 */
            double d = Double.parseDouble(s);
  …

Accepted Answer

All of this code is legal, and line 8 creates a new String with a value of "42.5". Lines 9 and 10 convert the String to a double and then back again. Line 11 is fun— Math.ceil()'s argument expression is evaluated first. We invoke the valueOf() method that returns an anonymous Double object (with a value of 42.5). Then the doubleValue() method is called (invoked on the newly created Double object), and returns a double primitive (there and back again), with a value of (you guessed it) 42.5. The ceil() method converts this to 43.0, which is cast to an int and assigned to x.

Question 5

What will be the output of the program? String s = "ABC"; 
s.toLowerCase(); 
s += "def"; 
System.out.println(s);

Accepted Answer

String objects are immutable. The object s above is set to "ABC". Now ask yourself if this object is changed and if so where - remember strings are immutable. Line 2 returns a string object but does not change the originag string object s, so after line 2 s is still "ABC". So what's happening on line 3? Java will treat line 3 like the following: s = new StringBuffer().append(s).append("def").toString(); This effectively creates a new String object and stores its reference in the variable s, the old String object containing "ABC" is no longer referenced by a live thread and becomes available for garbage collection.

Question 6

What will be the output of the program? public class ExamQuestion6 
{
    static int x; 
    boolean catch()
    {
        x++; 
        return true; 
    } 
    public static void main(String[] args)
    {
        x=0; 
        if ((catc…

Accepted Answer

Initially this looks like a question about the logical and logical shortcut operators "|" and "||" but on closer inspection it should be noticed that the name of the boolean method in this code is "catch". "catch" is a reserved keyword in the Java language and cannot be used as a method name. Hence Compilation will fail.

Question 7

What will be the output of the program? try 
{
    Float f1 = new Float("3.0");
    int x = f1.intValue();
    byte b = f1.byteValue();
    double d = f1.doubleValue();
    System.out.println(x + b + d);
}
catch (NumberFormatException e) /* Line 9 *…

Accepted Answer

The xxxValue() methods convert any numeric wrapper object's value to any primitive type. When narrowing is necessary, significant bits are dropped and the results are difficult to calculate.

Java.lang Class Questions