Exception handdling

import java.util.Scanner;

public class Main {

public static void main(String[] args) {

Scanner obj = new Scanner(System.in);

System.out.println("Enter your first Number");
int a = obj.nextInt();
System.out.println("Enter Your Second Number");
int b = obj.nextInt();
int sum = a / b;
System.out.println(sum);
throws BadNumberException{
if(b==0){
throws new BadNumberException("cannot divide by zero");

}


}


}
}
in the above code the bold text are wrong how to correct them

It's 'throw' not 'throws'; check any Java text book for the syntax.

kind regards,

Jos

spoon-feeding )

Code:

---

import java.util.Scanner;

class BadNumberException extends Exception {
    public BadNumberException(String message) {
        super(message);
    }
}
public class Main {

    public static void main(String[] args) throws Exception {
        Scanner obj = new Scanner(System.in);
        System.out.println("Enter your first Number");
        int a = obj.nextInt();
        System.out.println("Enter Your Second Number");
        int b = obj.nextInt();
        if(b==0) {
            throw new BadNumberException("cannot divide by zero");
        }
        int sum = a / b;
        System.out.println(sum);
    }
}

---

Hi
First of all your syntax is wrong for throws it is as follows
Syntax: return-type method-name(parameters-list) throws exception-list{
body of method
}

Complete and Running code is as follows
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import java.util.Scanner;

class BadNumberException extends Exception{
public BadNumberException(String message){
super(message);
}
}

public class Main {
public static void main(String[] args) {
try{
Scanner obj = new Scanner(System.in);
System.out.println("Enter your first Number");
int a = obj.nextInt();
System.out.println("Enter Your Second Number");
int b = obj.nextInt();
if(b==0){
throw new BadNumberException("cannot divide by zero");
}
else{
int sum = a / b;

System.out.println("Result is : "+sum);
}
}catch(BadNumberException e){
System.out.println(e.getMessage());
}

}

}
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