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Home/ Questions/Q 8800091
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T00:29:42+00:00 2026-06-14T00:29:42+00:00

I have the following code: Scanner inputSide = new Scanner(System.in); double side[] = new

  • 0

I have the following code:

Scanner inputSide = new Scanner(System.in);

double side[] = new double[3];
int i = 0;
do{
    try{
    System.out.println("Enter three side lengths for a triangle (each followed by pressing enter):");
    side[i] = inputSide.nextDouble();
    i++;
    }
    catch(Exception wrongType){
        System.err.println(wrongType);
        System.out.println("Please enter a number.  Start again!!");
        i=0;
    }
}
while(i<3);

It works fine and does what it’s meant to if I don’t enter a wrong data type but if I enter something other than a double then it loops over and over, printing everything in both try and catch blocks instead of waiting for me to enter another double.

Any help as to why it’s doing this – as I can’t seem to understand why – would be appreciated.

Thank you 🙂

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T00:29:43+00:00Added an answer on June 14, 2026 at 12:29 am

    The problem is that, you have used input.nextDouble method, which reads only the next token in the input, thus skipping the newline at the end. See Scanner.nextDouble

    Now, if you enter wrong value first time, then it will consider the newline as the next input. Which will also be invalid.

    You can add an empty input.nextLine in the catch block.

    catch(Exception wrongType){
        System.err.println(wrongType);
        System.out.println("Please enter a number.  Start again!!");
        i=0;
        input.nextLine();  // So that it consumes the newline left over
    }
    

    Now, your nextLine() will read the linefeed left over, and linefeed will not be taken as input to your nextDouble next time. In which case, it will fail, even before you giving any input.

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