using string to break out of loop in java

Question

I simply would like to break out of a while true loop if the user enters in "STOP" into the scanner. Currently my scanner only takes in integers and I believe this is where the problem lies. If I try to type in "STOP" I get many errors saying "exception in thread main". Here is a snippet of my code:

public class algorithm {
private Scanner input = new Scanner(System.in);
int n;

while (true){
    System.out.println("Eneter a number to check if it is odd or even, then hit enter: ");
    System.out.println("Type 'STOP' to end the program");
    n = input.nextInt();

    String strN = String.valueOf(n); //convert integer to string
    if (new String(strN).equals("STOP")){ //if string is STOP, breaks
        System.out.println("Thanks for using my program");
        break;
    }

    if (n % 2 == 0){
        System.out.println(n+" is even");
        continue;
    }
    else{
        System.out.println(n+" is odd");
        continue;

I know I am missing some closing curly braces but rest assured they are all there in my actual code. Thanks.

Here is the error I am getting: Exception in thread "main" java.util.InputMismatchException at java.util.Scanner.throwFor(Scanner.java:864) at java.util.Scanner.next(Scanner.java:1485) at java.util.Scanner.nextInt(Scanner.java:2117) at java.util.Scanner.nextInt(Scanner.java:2076) at OddOrEven.algorithm.checker(algorithm.java:13) at OddOrEven.main.main(main.java:7)

.nextInt()? you need to prompt for a string, then, not an int. — Marc B
– Marc B, Commented Jun 2, 2015 at 21:47
I don't think nextInt() will check if a user enters "stop". — Forseth11
– Forseth11, Commented Jun 2, 2015 at 21:47
Yes, I tried this already. But wouldnt I have to convert the strings into integers to check if they're odd or even? — Nicholas Roberts
– Nicholas Roberts, Commented Jun 2, 2015 at 21:48
Prompt for a string, then check whether that string is "STOP" or try to cast it to int otherwise. — tobias_k
– tobias_k, Commented Jun 2, 2015 at 21:48

Community · Accepted Answer · 2017-05-23 11:51:08Z

3

You've already identified the problem yourself - your Scanner reads only integers:

int n;
...
n = input.nextInt();

so it's impossible for the variable n (an int) to contain the string "STOP" (and your Scanner throws an exception anyway when you call nextInt() but it encounters a string, such as "STOP", that it cannot convert to an int).

To do this, you need to read strings from the input (probably using Scanner.nextLine()), check whether they are "STOP", and if not, only then attempt to convert them to ints using something like:

int n = Integer.parseInt(mystring)

To handle garbage input (neither STOP nor an integer), wrap the parseInt line in a try-catch block so that you can detect when the input is garbage by catching the Exception

try {
  int i = Integer.parseInt(mystring);
  // do something with the int
}
catch (NumberFormatException e) {
  // display warning message to the user instead
}

3 Comments

Nicholas Roberts Over a year ago

How could I check if the user typed in a garbage string, such as "dsfasdfsa" and then print out something like "Please enter a valid input"? Cause at the moment, I am doing exactly what you did, if the user didnt type in STOP, i cast it to a string. But I get many errors if I type in a garbage string.

DNA Over a year ago

Wrap the parseInt line in a try-catch block so that you can detect when the input is garbage by catching the Exception

Nicholas Roberts Over a year ago

Hey, I actually created a separate method that contained a try-catch block, Worked great. Thank you so much!

GregH · Accepted Answer · 2015-06-02 21:57:00Z

1

The most simple way is probably to use input.next() instead of input.nextInt(). Using input.next() will read input in as a String, then you can check if the input is equal to "QUIT", if it is not you can use Integer.parseInt to parse the Integer from the read string

answered Jun 2, 2015 at 21:57

GregH

5,53112 gold badges62 silver badges115 bronze badges

Comments

Polynomial Proton · Accepted Answer · 2015-06-02 21:57:55Z

Something like below, should work out.

NOTE: havent tested compile errors, just wrote it out(but you get a gist)

     public Scanner input = new Scanner(System.in);
    int n;

while (true){
    System.out.println("Eneter a number to check if it is odd or even, then hit enter: ");
    System.out.println("Type 'STOP' to end the program");
    n = input.next();
    Integer input;

   // String strN = String.valueOf(n); //convert integer to string
    if (strN.equals("STOP")){ //if string is STOP, breaks
        System.out.println("Thanks for using my program");
        break;
    }
    else{
        try{
    input=  Integer.parseInt(strN);

        }
        catch(Exception e)
        {
            System.out.println("Please enter a  number");
        }
    }

    if (input % 2 == 0){
        System.out.println(n+" is even");
        continue;
    }
    else{
        System.out.println(n+" is odd");
        continue;
    }

Collectives™ on Stack Overflow

using string to break out of loop in java

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related