🚀 go-pugleaf

One solution is to use an array of size 1, arrays are are changed and
not copied.. working program this time ..
Ill try to get a better solution and get back to u...

class Hej {
	Hej() {
	int[] a = new int[1];
	a[0]=0;
	System.out.println(a[0]);
	change(a);
	System.out.println(a[0]);
	}
	public void change(int b[]) {
		b[0]=7;
	}
	public static void main(String args[]) {
		Hej newProg = new Hej();
	}
}



On Tue, 20 Mar 2001 11:05:29 +0100, "Nikola Mitrovic"
<nikola@mail.com> wrote:

>Nope, won't work;
>
>First, there is no method set, and there is no method .valueOf(int), so I
>guess you must use .decode(String), or .parseInt(String). But guess, it
>wont'k work then either!
>
>Also, the same thing with String class doesn't work, and String is an
>object, and can be easiliy assigned, concatenated, etc...
>
>what I did more is to say
>
>        test (Integer param) {
>                Integer param2 = new Integer(5);
>                param = param2;
>        }
>
>but it didn't work either, and it should since, it should be pointer
>assignment...
>
>plz, any working example?
>
>"Stefan Adolfsson" <stefan.adolfsson@usa.net> wrote in message
>news:3ab688c8.43123726@news.cs.umu.se...
>> Use the integer as an object,,,
>> instead of:
>> int param ;
>> do:
>> Integer param;
>> then the integer will work as an object which is changed directly in
>> java and not copied.
>> example:
>>
>> public void test(Integer param) { param.set(5);
>> }
>>
>> public void example(){
>>         Integer example_parameter = new Integer(0);
>>         test( example_parameter );
>>         System.out.println("parameter: " + example_parameter);
>> }
>>
>>
>>
>>
>>
>> On Mon, 19 Mar 2001 21:29:30 +0100, "John Good"
>> <dumster@collector.org> wrote:
>>
>> >
>> >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"...
>> >
>> >plz help :)
>> >
>> >
>>
>
>