Can you cast a method argument?

[David Junta]

I have a method:
public static void methodName( long x , int y )
If I want to use this method but some of my y's are longs, can i do this in those cases:
methodName( sampleInteger , ( int )( sampleLong ) );
I got a compile error but I'm wondering if there is another way to do this. If not, do you have to make a new, slightly different method for these cases?
Any help would be much appreciated

[Steve Fahlbusch]

Is the below an example of what you are discussing? If so
it works fine.

class TestCase {
   public static long Case ( long a, int b ) {
      return a + b;
   }
   public static void main ( String[] args ) {
   
      int ma = 3;
      long mb = 4;
      System.out.println("Case ( 3, 4 ) = " + Case( ma, (int)mb));
   
   }
}

[John Bateman]

Originally posted by David Junta:
If I want to use this method but some of my y's are longs, can i do this in those cases:
methodName( sampleInteger , ( int )( sampleLong ) );

Hi
IMHO if some of your "y's" are long, then make the parameter long. Then just cast the result to an int if and when you need it. This will guarantee you don't 'lose' information when converting a potentially long 'long' into an int.

public long methodName(long x, long y) {
  // performn everything as a long
  // convert the result to an int OUTSIDE the method 
  // if/when you need an int.
}

Since an int is 32 bit and a long is 64 you could run the risk of losing information when you perform a 'narrowing cast' (64 bits pushed into 32 bits), or, casting your long variable to an int in the method call. But if you have a 32 bit (int) and cast it into a 64 bit (long) you NEVER will lose any information.
Hope this helps.

[This message has been edited by John Bateman (edited March 15, 2001).]