Array reference

Question

If I create an array of variables:

public String a;
public String b;
public String c;

public String[] d = {a,b,c};

it will contain {null, null, null};

If I then do this:

a = "5";
System.out.println(d[0]);

output will be null since a was null when d was initialised.

Is there a way to create an "array of references", so that output in this case would be 5?

EDIT: PLOT TWIST!

First part of code is in some class. Second part is in a class that is extending the first one.

How are you getting null from primitives? Also, those variables haven't been initialized; the code won't compile. — Makoto
– Makoto, Commented Nov 23, 2013 at 19:19
"it will contain {null, null, null};" No your code won't compile. — Alexis C.
– Alexis C., Commented Nov 23, 2013 at 19:19
Output will not be null, it will be zero (0), because primitives integers are initialized to zero. Use the object wrappers (Integer). — Paul Richter
– Paul Richter, Commented Nov 23, 2013 at 19:20
OKay, my bad, I tried to simplify it too much. I was using Strings... — Karlovsky120
– Karlovsky120, Commented Nov 23, 2013 at 19:21

arshajii · Accepted Answer · 2013-11-23 20:12:14Z

1

You can wrap the Strings with your own mutable class:

class MyString {
    private String value;

    public MyString(String value) {
        this.value = value;
    }

    void setValue(String value) {
        this.value = value;
    }

    @Override
    public String toString() {
       return value;
    }
}

Now:

public MyString a = new MyString(null);
public MyString b = new MyString(null);
public MyString c = new MyString(null);

public MyString[] d = {a,b,c};

...

a.setValue("5");  // <--

You might want to ensure that a is not null before calling setValue(), however.

edited Nov 23, 2013 at 20:12

answered Nov 23, 2013 at 19:29

arshajii

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2 Comments

arshajii Over a year ago

@MarkoTopolnik Why do they need explicit initialization? They're fields, not local variables.

Marko Topolnik Over a year ago

Because a.setValue("5") results in an NPE as you have showed it, and d is still {null, null, null}.

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