CODEWITHSUNDEEP
javaloopsinputcharacter
Francis Putnam
Francis Putnam

Reputation: 33

Breaking out of an infinite loop after a user inputs a character

I want to make an infinite loop of a message and stops it if a user inputs a character for example "x". But its getting an error on the if statement saying that it is an unreachable code. Can someone help me.

import java.util.Scanner;
public class Loop {

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

        while (true) {
            System.out.println("Good Morning!!!");
        }
        y = input.next().charAt(0);
        if (y == 'y') {
            break;
        }
    }   
}

Upvotes: 0

Views: 674

Answers (4)

Utkarsh Sahu
Utkarsh Sahu

Reputation: 419

Yes, the code which you have put is actually unreachable. The while() loop will go on executing continuously since the if statement is outside the while() block. You can put the if statement inside the while block like this.

import java.util.Scanner;
public class Loop {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        char y;
        while (true) {
            System.out.println("Good Morning!!!");
            y = input.next().charAt(0);
            if (y == 'y') {
                break;
            }
        }
    }   
}

Upvotes: 0

user17233545
user17233545

Reputation:

Try this.

public static void main(String[] args) throws IOException {
    InputStream input = System.in;
    while (true) {
        System.out.println("Good Morning!!!");
        if (input.available() > 0 && input.read() == 'y')
            break;
    }
}

Upvotes: 0

ʀᴀʜɪʟ
ʀᴀʜɪʟ

Reputation: 298

Acknowledgement

A thread is a thread of execution in a program. The Java Virtual Machine allows an application to have multiple threads of execution running concurrently. Every thread has a priority. Threads with higher priority are executed in preference to threads with lower priority. Each thread may or may not also be marked as a daemon. When code running in some thread creates a new Thread object, the new thread has its priority initially set equal to the priority of the creating thread, and is a daemon thread if and only if the creating thread is a daemon.

When a Java Virtual Machine starts up, there is usually a single non-daemon thread (which typically calls the method named main of some designated class). Read more about threads here

Solutions

Because both input & loop blocks the thread, you have to put them in seperate thread. Just like this :-

public class Demo {
    volatile static char input;

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

        new Thread(new Runnable() {
            @Override
            public void run() {
                while (input != 'y'){
                    System.out.println("Good morning !");
                }
            }
        }).start();

        input = scanner.next().charAt(0);
    }
}

This will print "Good morning !" until the input != 'y'. On the other hand, we are reading the input from user. As user enteres 'y', the condition get false and loop stops.

Upvotes: 0

Erik McKelvey
Erik McKelvey

Reputation: 1627

Your if statement is not inside the while loop:

import java.util.Scanner;
public class Loop {

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

        while (true) {
            System.out.println("Good Morning!!!");
            y = input.next().charAt(0);
            if (y == 'y') {
                break;
            }
        }
    }   
}

Upvotes: 1

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