Switch statement within Do-While loop doesn't exit

Question

The code doesn't exit after I type "stop" - for some reason. Why? Step-by-step debugging shows that after I enter "stop" it's value consists of exactly 's','t','o','p' without any line breaks, etc. - however, the code still goesn't exit. Could anyone tell why, please?

import java.util.Scanner;

public class Application {
public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    // asking username
    System.out.print("Username: ");
    String username = input.nextLine();

    String inpText;
    do {
        System.out.print(username + "$: ");
        inpText = input.nextLine();
        System.out.print("\n");
        // analyzing
        switch (inpText) {
        case "start":
            System.out.println("> Machine started!");
            break;
        case "stop":
            System.out.println("> Stopped!");
            break;
        default:
            System.out.println("> Command not recognized");
        }
    } while (inpText != "stop");

    System.out.println("Bye!..");
}
}

read this: How do I compare strings in Java? and then fix your while condition. — jlordo
– jlordo, Commented Aug 2, 2013 at 10:27
Thanks @jlordo. Couldn't imagine the reason was in the comparison itself; and the linked question covers everything. — evictorov
– evictorov, Commented Aug 2, 2013 at 14:20

Community · Accepted Answer · 2017-05-23 10:33:30Z

2

To compare Strings use .equals() and not ==, unless you really know what you are doing.

inpText != "stop" //Not recommended
!"stop".equals(inpText) //recommended

You cannot use a String in a switch unless you are using jdk 1.7+.

Cannot switch on a value of type String for source level below 1.7. Only convertible int values or enum variables are permitted

edited May 23, 2017 at 10:33

CommunityBot

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answered Aug 2, 2013 at 10:42

rocketboy

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2 Comments

evictorov Over a year ago

Thanks @rocketboy. JDK is 7u17, and looks like it was a matter of correct comparison. However, acc. to specification of the equals() shouldn't !inpText.equals("stop") be more correct?

rocketboy Over a year ago

String equals is symmetric, so a.equals(b) is same as b.equals(a). That said, if inpText = null, inpText.equals("stop") can throw a NPE but it won't be thrown for "stop".equals(inpText)

Tristan Van Poucke · Accepted Answer · 2013-08-02 10:33:32Z

0

You are comparing pointers not strings with this piece of code:

while (inpText != "stop");

Should be something like this:

while (!"stop".equals(inpText));

answered Aug 2, 2013 at 10:33

Tristan Van Poucke

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2 Comments

evictorov Over a year ago

Thanks Tristan. However, according to the function' specification shouldn't !inpText.equals("stop") be more correct?

Tristan Van Poucke Over a year ago

evictorov: by writing it as !"stop".equals(inpText) you easily avoid any possible nullpointers. Its just a little trick I use quite often.

Ilesh Patel · Accepted Answer · 2013-08-02 10:34:10Z

0

change while (inpText != "stop"); to while (!(inpText.equals("stop")));

answered Aug 2, 2013 at 10:34

Ilesh Patel

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Comments

Mohsin Shaikh · Accepted Answer · 2013-08-02 11:33:45Z

0

If your JDK is 1.6 or lower you can't switch() on a String

P.S. switching on a String is probably not the best solution Yeah in java 1.6 you can only switch int, boolean, double, long, and float I believe.

answered Aug 2, 2013 at 11:33

Mohsin Shaikh

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1 Comment

evictorov Over a year ago

Thanks Mohsin. Mine is 7u17.

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