Question

Could anyone please explain to me why the add() method returns 0 instead 4?
I am trying to use int 0 in case “invalid” string number is provided (e.g. four).
I am getting the right results for String arguments 3,4 / 3,four / three,four / but not for three, 4.

Can you please give me hints what I am doing wrong?
Thanks!

public class Numbers2
{
    public static void main(String[] args) {

        System.out.println(add("d","4"));
    } // main() method

   public static int add(String a, String b)
   {
       int x = 0;
       int y = 0;

      try{ 
          x = Integer.parseInt(a); 
          y = Integer.parseInt(b);

          System.out.println("No exception: " + (x+y));

          return x + y;
      }
      catch (NumberFormatException e){

          if(x != (int) x ){              
              x = 0;
          }
          if(y != (int) x ){
              y = 0;              
          }      
          return x + y;  
      }

   } // add() method
} // class

Answer 1

The problem is that the line y = Integer.parseInt(b); doesn’t get the chance to execute as you are passing "d" as the first argument which is not an integer, the line x = Integer.parseInt(a); results in exception and both x and y remain 0.

You could solve the problem by using separate try/catch for both:

int x = 0; //x is 0 by default
int y = 0; //y is 0 by default

try { 
    x = Integer.parseInt(a); //x will remain 0 in case of exception  
}
catch (NumberFormatException e) {
    e.printStackTrace();
}

try {  
    y = Integer.parseInt(b); //y will remain 0 in case of exception  
}
catch(NumberFormatException e) {
      e.printStackTrace();
}

return x + y; //return the sum