How to retry parsing a double when a NumberFormatException is thrown?

Question

If a NumberFormatException is thrown when I parse a double from a string (given by the user), how can I retry?

String input = JOptionPane.showInputDialog(null, message + count);
double inputInteger = Double.parseDouble(input);

See http://stackoverflow.com/questions/8391979/does-java-have-a-int-tryparse-that-doesnt-throw-an-exception-for-bad-data — Tim S., Jun 16 '13 at 12:28

score 0 · Answer 1 · answered Jun 16 '13 at 12:29

0

inputInteger = null;

while(inputInteger == null)
{
    input = JOptionPane.showInputDialog(null, message + count);
    try
    {
        if (isValidGrade(input, maxPoints))
            inputInteger = Double.parseDouble(input);
    }
    catch(NumberFormatException e)
    {
        // Show your error here
        inputInteger = null;
    }
}

answered Jun 16 '13 at 12:29

Devolus

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if `inputInteger` is of type `double` this will not work, since it requires `Double`. – Micha Wiedenmann Jun 16 '13 at 12:31
@MichaWiedenmann _if `inputInteger` is of type `double`_ Would you say integer ? – Alexis C. Jun 16 '13 at 12:40
I have no idea, but the function called is most certainly `Double.parseDouble` and it is very likely that the variable was of type `double`, but you have to ask the OP. – Micha Wiedenmann Jun 16 '13 at 12:59

score 0 · Answer 2 · answered Jun 16 '13 at 12:29

0

You need to wrap your code in try catch and handle the exception to do whatever you want. Do sth like this:

       try {
                inputInteger = Double.parseDouble(input);
       }catch(NumberFormatException nfe) {
           // Go to take input again
       }

answered Jun 16 '13 at 12:29

Juned Ahsan

67,789
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*// Go to take input again* :Please explain how will he do that from inside the catch ? – AllTooSir Jun 16 '13 at 12:31
1

Idea is not to spoon feed people here. He needs to understand how to catch exception and work accordingly. We are not here to code the entire thing here. – Juned Ahsan Jun 16 '13 at 12:33
*Idea is not to spoon feed people here. He needs to understand how to catch exception and work accordingly.* He knows he is getting an exception , he wanted to know how to get the i/p again ! – AllTooSir Jun 16 '13 at 12:33
Well then just leave a comment instead of an answer which is not an answer. – Micha Wiedenmann Jun 16 '13 at 12:33

score 0 · Answer 3 · answered Jun 16 '13 at 12:43

0

You can do this by calling the method recursively in the catch block. For an example:

public void yourMethod() {
     try {
          input = JOptionPane.showInputDialog(null, message + count);
          if (isValidGrade(input, maxPoints)){
          inputInteger = Double.parseDouble(input);
    } catch (NumberFormatException e) {
          this.yourMethod();
    }
}

This is not a working code but use the concept in your code. Using a while loop is also another option. But I prefer this method over a while loop because this reduces the memory overhead.

answered Jun 16 '13 at 12:43

Sithum

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7

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"reduces the memory overhead"? Actually, this uses *more* memory than a loop would. – Ryan Stewart Jun 16 '13 at 12:55
This can go into infinite loop!! – Lokesh Jun 16 '13 at 13:54
Yes, my bad. Actually it doesn't reduces the memory overhead. It seems I had a totally wrong understanding of the this concept. I should read more about it. Thanks for the comment Stewart. – Sithum Jun 16 '13 at 14:05

score 0 · Answer 4 · answered Jun 16 '13 at 12:50

You can use a do-while loop to repeat the things in this case and make boolean variable false when all conditions are met.

Here is an example:

    boolean isFailure=true;
     do{
        try{
            input = JOptionPane.showInputDialog(null, message + count);
            // do whatever you want....here...
            isFailure=false;

       }catch(NumerFormatException e){
            //log the exception and report the error
            JOptionPane.showMessageDialog(null,"Invalid Input! Try again!", "Error", 
 JOptionPane.ERROR_MESSAGE); 

         }

         }while(isFailure);

How to retry parsing a double when a NumberFormatException is thrown?

4 Answers4