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Static method and variables1796

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Static method and variables1796
What will be the output of the following program?
public class Vinayak {
    static int laddu = 4;
    final static String TRUTH = "GaneSh iS soO great";
    static String trunk = "";
    static float mouse = 12.0f;
    public static void main(String[] args) {
    public static void ganesh() {
        switch (laddu) {
            case 5 * 3 :
            case 18 :
                laddu -= 3;
            case ((int) mouse) :
                laddu = 2;
            default :
                laddu = TRUTH.length() - 1;
            case 2 :
                laddu = 0;
    public static void vignesh(int i) {
        trunk += TRUTH.charAt(i);

A. StrOnG
B. Str nG
C. strong
D. Some other output
E. Compilation Error or Runtime Error
Topic: charAt() Method In Java - Java Character Extraction

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User comments below. All of them might not be correct.

Ans is e:
because constant expression is required in switch case and (int)mouse is not constant expression

Posted by Mânïshå Mùlchåndânï    2014-07-16 13:12:56

ans e.compilatn err
case((int)mouse) err
case expressn must be constant

Posted by Maheshwari Natarajan    2014-07-16 13:13:24

Answer : A
StrOng will get printed as the length of the string is being manipulated and characters at diff location inthe String TRUTH are being accessed

Posted by Deepak Kumar Yadav    2014-07-16 13:13:30

Answer E.
switch case ly accept constant expression.
compilation error

Posted by NagaRaj Nataraj    2014-07-16 13:18:38

answer is E becoz case expresions must be constants only but in case (int)mouse) it is not a constant.

Posted by Vijay Kumar    2014-07-16 13:21:20

Congratulations Maheshwari Natarajan. You are this dose winner. We will send you the link using which you can claim your recharge.

Posted by Merit Campus    2014-07-17 09:13:07

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