🚀 go-pugleaf

In article <Z_tt6.19$xH3.1481@psinet-eu-nl>, John Good
<dumster@collector.org> writes
>I've got a strange problem with Java, it's probably a feature, but please
>point me to solving this.
>
>If I have something like:
>
>public void test(int param) { param = 5;
>}
>
>and practically want to set the param to 5 (I do not want to use return,
>please!), and when I call it from another method, like:
>
>public void example(){
>        int example_parameter = 0;
>        test( example_parameter );
>        System.out.println("parameter: " + example_parameter);
>}
>
>I'll get the 0 as the output, and I want 5; how to assign a value a
>parameter? (pointers I miss you ;)
>I've checked some books on that, but as allways its not in it, because it's
>"too obvious"...


 Wow; even in C++ what you have written would not work without
 declaring   public void test( int &param ){.........
 That little "&" makes a difference :-> !

 In Java variables EXIST as one-word values for a simple numerical
 type and EXIST as one-word locations for class or array objects.
 If you pass an integer, you pass the numerical value.
 If you pass a class-object, you pas the location of the object
 as allocated in the calling routine.  

 This will work:

 public class WrapsInt{
    public   int x;
    WrapsInt(){      x=0; };
    WrapsInt(int h){ x=h; }
 }
 public class SomeWork{
    ...............
    public void doFirst(){
       WrapsInt     exampleParam = new WrapsInt(0);
       doInside(    exampleParam );
       int result = exampleParam.x; ........
    };
    public void doInside( WrapsInt theParam ){
       theParam.x = 5;
       .......
    };
    ........
 }

 'result' will be five, not zero.

 It's not really a lot of work because usually, when you want to
 change (some fields of) what you passed as parameter, what you
 passed will be a class object anyway; not a simple numerical type. 



 [As best I remember class "Integer" doesn't do what you want,
 it has a single int field which has  **constant** value].



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