Answers for "java Throwable, Error und Exception"

2

java throw an exception

public static void main(String[] args) {
	Scanner kb = new Scanner(System.in);
    System.out.println("Enter a number");
    try {
    	double nb1 = kb.nextDouble();
    	if(nb1<0)
        	throw new ArithmeticException();
        else System.out.println( "result : " + Math.sqrt(nb1) );
    } catch (ArithmeticException e) {
        System.out.println("You tried an impossible sqrt");
    }
}
Posted by: Guest on October-16-2020

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