Pass function argument to callback function in Java

Question

I have this code:

public void FbShare(JSONObject parameters) {

    if (Session.getActiveSession() == null || !Session.getActiveSession().isOpened()) {

        Session.StatusCallback callback = new Session.StatusCallback() {

            @Override
            public void call(Session session, SessionState state, Exception exception) {

                if (state.isOpened()) {
                    publishFeedDialog(parameters);  // <--- HERE
                }
            }
        };

        Session.openActiveSession(this, true, callback);
    } else {
        publishFeedDialog(parameters);
    }
}

Why is parameters not accessible in the publishFeedDialog(parameters)call?

make it final -> public void FbShare(final JSONObject parameters) — A4L
– A4L, Commented Oct 15, 2014 at 14:10

bruno · Accepted Answer · 2018-05-18 15:47:28Z

4

Set the parameters parameter as final:

public void FbShare(final JSONObject parameters) {
 //...    
}

For an overview of what's happening, see:

edited May 18, 2018 at 15:47

answered Oct 15, 2014 at 14:10

bruno

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Comments

duffymo · Accepted Answer · 2014-10-15 14:09:52Z

4

Trying making that argument final:

public void FbShare(final JSONObject parameters) {

answered Oct 15, 2014 at 14:09

duffymo

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4 Comments

Narek Over a year ago

final in java is the same as const in C++? Does it mean that I cannot change the values of parameters argument?

duffymo Over a year ago

No, not the same. final says the reference won't change. Java is pass by value, and for objects the value is a reference type. You can't change the reference, but you can change the mutable state of the object the reference points to if it allows you to. const in C++ is a much wider ranging keyword.

Narek Over a year ago

You mean the reference cannot point to other object, or the object it is pointing will not be changes inside the function.

duffymo Over a year ago

I think I was clear: the reference itself cannot change.

Tassos Bassoukos · Accepted Answer · 2014-10-15 14:10:10Z

2

Change the first line to

public void FbShare(final JSONObject parameters) {

answered Oct 15, 2014 at 14:10

Tassos Bassoukos

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Comments

slim · Accepted Answer · 2014-10-15 14:28:23Z

The compiler error message tells you everything you need to know, if you can decipher it:

 Cannot refer to the non-final local variable parameters in an enclosing scope

That is, the variable parameters in the enclosing scope is not final, so it can't be referred to.

To fix that, make it final:

 public void FbShare(final JSONObject parameters) {

Java isn't clever enough to keep track of what parameters refers to, unless you use final to guarantee that it's always going to point at the same object.

This would equally apply if you were referring to a local variable that wasn't a method parameter:

 public void foo() {
      final String bar = ...;
      Callback callback = new Callback() {
           void invoke() { 
               something(bar); 
           }
      };

SMA · Accepted Answer · 2014-10-15 14:12:25Z

0

Make your parameters as final so it can be used within your anonymous class.

answered Oct 15, 2014 at 14:12

SMA

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