Why doesn't this code terminate my program? Java

Question

I am trying to implement the System.exit(0); in java to terminate my program when the word "exit" is typed into the console. I wrote the following method:

    public static void exit(){

    Scanner input = new Scanner(System.in);
    Str1 = input.next(String);

    if (str1 = "exit"){

        System.exit(0);
    }

    else if (str1 = "clear"){

        System.out.println("0.0");
    }       
}

and it doesn't seem to be working. Does anyone have any suggestions?

Thanks P.S the "clear" is just supposed to return 0.0 when "clear" is entered into the console, if you could not already tell.

If you would have googled how to compare strings you would have got your answer. :) — NewUser, Oct 11 '13 at 07:28
btw besides equals(), I think your codes got some minor mistakes. Str1 and str1 are different. — gjman2, Oct 11 '13 at 07:44

score 4 · Answer 1 · edited May 23 '17 at 11:53

4

Compare strings with equals() not with ==.

The reason is that == just compares object references/primitives,where as String's .equals() method checks equality.

if (str1.equals("exit")){

}

and also

else if (str1.equals("clear")){

}

Might useful :What are the benefits of "String".equals(otherString)

edited May 23 '17 at 11:53

Community

1
1

answered Oct 11 '13 at 07:25

Suresh Atta

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better - `"exit".equals(str1)` – sanbhat Oct 11 '13 at 07:28

score 1 · Answer 2 · answered Oct 11 '13 at 07:25

1

if(str.equals("exit"))

or

if(str.equalsIgnoreCase("exit"))

or

if(str == "exit")

instead of

if (str1 = "exit"){

answered Oct 11 '13 at 07:25

newuser

8,338
2
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score 1 · Answer 3 · edited Oct 23 '13 at 13:19

1

With if (str1 = "exit") you use an allocation instead of a compare. You can compare with the equals() method.

edited Oct 23 '13 at 13:19

Andrew Barber

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answered Oct 11 '13 at 07:43

JulianG

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score 0 · Answer 4 · answered Oct 11 '13 at 07:30

0

Use the String.equals(String other) function to compare strings, not the == operator.

The function checks the actual contents of the string, the == operator checks whether the references to the objects are equal. Note that string constants are usually "interned" such that two constants with the same value can actually be compared with ==, but it's better not to rely on that.

So use:

if ("exit".equals(str1)){

}

answered Oct 11 '13 at 07:30

code_fish

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These were all very helpful, thank you. However, I implemented these changes and the console is still throwing an exception rather than terminating the program. – Charlie Tidmarsh Oct 11 '13 at 07:38
Throwing what exception ? @CharlieTidmarsh – gjman2 Oct 11 '13 at 07:40
@CharlieTidmarsh Could you tell me your exception? – code_fish Oct 11 '13 at 07:47

gjman2 · Answer 5 · 2013-10-11T08:16:01.373

0

Besides equals(), the input.next(String pattern); require pattern not String data type

Change your code to:

public static void exit(){   

Scanner input = new Scanner(System.in);
str1 = input.next(); //assumed str1 is global variable

if (str1.equals("exit")){

    System.exit(0);
}

else if (str1.equals("clear")){

    System.out.println("0.0");
}

}

Notes : http://www.tutorialspoint.com/java/util/scanner_next_string.htm

edited Oct 11 '13 at 08:16

answered Oct 11 '13 at 07:46

gjman2

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17
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2

Change `str1 = input.next();` to `String str1 = input.next();`, except if `str1` is a global variable – JulianG Oct 11 '13 at 07:51

Why doesn't this code terminate my program? Java

5 Answers5