Why does my error my exceptions handling cause an infinite loop?

Question

I'm having trouble with my exception handling. The program runs fine if I input a number but create an infinite loop if a character is entered.

boolean ask= true;
    while(ask)
    {
        ask = false;
        try 
        {
            System.out.println("What is the age?");
            int age = input.nextInt();  
            setAge(age);
        }catch(InputMismatchException e) {
            System.out.println("Invalid input!");
            ask = true;
        }
    }//end while

Answer 1

Let's say you enter "abc"

Your call to input.nextInt() causes the scanner to look at the a and say "That's not an int, so I will throw an exception."

In the exception handler, you set ask to true so the loop repeats.

When the loop repeats, the scanner looks at that exact same a again, and it says "That's not an int, so I will throw an exception."

In the exception handler, you set ask to true so the loop repeats.

And so on....

That pesky a never gets consumed by the scanner.

Answer 2

Try below code:

      boolean ask= false;
        while(!ask)
        {
           try
           {
               System.out.println("What is the age?");
               int age = input.nextInt();//does not read the newline character in your input created by hitting "Enter,"
               setAge(age);
               ask = true;
           }catch(InputMismatchException e) {
               System.out.println("Invalid input!");
               input.nextLine();//consumes the \n character
           }
        }//end while

Answer 3

From the source code of nextInt:

    public int nextInt(int radix) {
        // Check cached result
        if ((typeCache != null) && (typeCache instanceof Integer)
            && this.radix == radix) {
            int val = ((Integer)typeCache).intValue();
            useTypeCache();
            return val;
        }
        setRadix(radix);
        clearCaches();
        // Search for next int
        try {
            String s = next(integerPattern());
            if (matcher.group(SIMPLE_GROUP_INDEX) == null)
                s = processIntegerToken(s);
            return Integer.parseInt(s, radix);
        } catch (NumberFormatException nfe) {
            position = matcher.start(); // don't skip bad token
            throw new InputMismatchException(nfe.getMessage());
        }
    }

It uses Integer.parseInt(s, radix); to produce the result.
If call Integer.parseInt("s"); will result:

Exception in thread "main" java.lang.NumberFormatException: For input string: "s"