SImple but important java question

Question

hello all,
just been revising for an exam and just have come across this bit where my notes are trying to explain that arguments passed to java methods are PASSED BY VALUE, and the example is this:
public class TestSwap {
   public static void swap( int x, int y )
      int temp;
      temp = x; x = y; y = temp;
      return;
   public static void main( String [] args )
      int a = 4;
      int b = 5;
      swap( a, b );
      System.out.println( "a=" + a ); // 4
      System.out.println( "b=" + b ); // 5
}Isn't this obvious since when a and b are printed out they are already 4 and 5 because of this:
      int a = 4;
      int b = 5;And the method swap(a,b) method might not just as well be there. because it won't have an affect on when a and b are pinted out???!!!!
Am I jumping the gun, if so can some please explain, i know its important but i cant see it!
Thanks :)

Answer 1

Ok, we are proving, got it.
What the, i cant work out how this works! The example
you gave?
Is it because of the arrays referencing
a[0]...a[1]..etc????!!!Note the difference between swapping a and b vs. swapping a[0] and b[0]. Also, if there were a member variable, swapping a.thing and b.thing would be reflected in the caller.
With your initial example, we're changing the values of the variables, and since the method and the caller each have their own copies of those variables, the caller is not affected. (Note that only a primitive, a reference or null, can be a variable's value in Java, never an object.)
With array elements and member variables of parameters, we're changing the contents of the objects pointed to by those variables. Since there's only a single object for each parameter, and the method's variable and caller's variable both point to that same object, both see the change.
References are passed by value, so changing the reference's value (which object it points to) doesn't affect the caller. But only the reference is copied, not the object, so changing the state of the object will be seen by the caller.