Why am I getting java.lang.ArithmeticException: / by zero?

Question

I am a beginner in java and my code to generate a random sums throws a strange exception...

    public void randomRekensom(int n)
{

    switch(n) {
        case 1: this.max = 100;
                break;
        case 2: this.max = 150;
                break;
        case 3: this.max = 200;
                break;
}
    getal1= (int) Math.sqrt(max);
    getal2= (int) Math.sqrt(max);

    operator=ThreadLocalRandom.current().nextInt(1, 4 + 1);
    switch(operator) {
        case 1: antwoord=(this.getal1+this.getal2);
                operatorTeken=" + ";
                break;
        case 2: antwoord=(this.getal1-this.getal2);
                operatorTeken=" - ";
                break;
        case 3: antwoord=(this.getal1/this.getal2);
                operatorTeken=" / ";
                break;
        case 4: antwoord=(this.getal1*this.getal2);
                operatorTeken=" * ";
                break;
}

}

Maybe it's because I've been staring too much at my screen today but I don't know why I'm getting this error.

Thanks in advance!

Thanks, I was derping too hard, it gets called from another class and your comment made me look through my code again, sorry for the simple question. — iSidle, Mar 29 '16 at 22:09

Andy Turner · Accepted Answer · 2016-03-29T22:17:46.753

You only set this.max if n is 1, 2 or 3. If you've not set it to some other value previously, this.max == 0, so getal2 == Math.sqrt(0) == 0.

You should add a default case to your switch statement to handle all other values of n. It may be appropriate simply to throw an IllegalArgumentException.

switch(n) {
    case 1: this.max = 100;
            break;
    case 2: this.max = 150;
            break;
    case 3: this.max = 200;
            break;
    default: throw new IllegalArgumentException("Not 1, 2 or 3");
}

or you may have a sensible default value to which you can set this.max.

Why am I getting java.lang.ArithmeticException: / by zero?

1 Answers1