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Java: Why Program Skip Char Login?

Java Skip First Character Input When does a program catch an exception?

System.out.println("Enter Character:");
f = s.next().charAt(0);

Program code:

    public class Main {
            public static void main(String[] args) {
             int num;
             char f = 'y';
              Scanner s = new Scanner(System.in);               

                do {
                    try {
                    System.out.print("Enter Number:");
                    num = s.nextInt();


                    catch (InputMismatchException e) {
                    System.out.println("False=> This is Not Integer");
                }
            System.out.println("Enter Character:");
            f = s.next().charAt(0);

            while(f != 'y' && f !='n') {
            System.out.println("Press 'y' or 'n'");
            f = s.next().charAt(0);
        }
        }
           while(f == 'y');
           System.out.print("Print:" + f);
       }
       }

Compiler output:

    Enter Number:ghjgh
    False=> This is Not Integer
    Enter Character:(Compiler Skip This Input)
    Press 'y' or 'n'
    n
    Print:n

Why is this happening. I don't know why, it is missing input in the catch catch.

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2 answers


Add s.nextLine()

after scanning the number



...
try {
    System.out.print("Enter Number:");
    num = s.nextInt();
}
catch (InputMismatchException e) {
    System.out.println("False=> This is Not Integer");
}
s.nextLine();
...
+1


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You can use the instanceof keyword, just like this:



try{
    String input = s.nextInt();
    int num = Integer.parseInt(input);
    // if execute here input is number
}catch(Exception e){
    // if execute here input is not number
    System.out.println("Please input number!");
}
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