[QFJ-613] Provide correct example in chapter "Creating Your Application" Created: 21/Jun/11  Updated: 04/Aug/11  Resolved: 04/Aug/11

Status: Closed
Project: QuickFIX/J
Component/s: Documentation
Affects Version/s: 1.5.0
Fix Version/s: 1.5.1

Type: Improvement Priority: Default
Reporter: dario marguez Assignee: Jörg Thönnes
Resolution: Fixed Votes: 0
Labels: None


 Description   

Hi, I'm following the instructions on the page Creating Your QuickFIX/J Application. It tells me to create an app by implementing the Application interface, but the code for MyClass is instantiating an object for the interface. I thought this is prohibited in Java?



 Comments   
Comment by Grant Birchmeier [ 21/Jun/11 ]

I guess technically he's right, though my impression is that the code is more of an example and not intended to actually compile.

The line:
> Application application = new Application();
probably could be changed to something like:
> Application application = new YourClassThatImplementsApplication();
which I think would clear up any confusion.

Dario, though there is a slight correction needed here, did you really create this bug because you had a question and could not find anywhere else to submit it? (Forgive me if I am mistaken.) You should really be joining the mailing list and submitting questions there. Bugs should only be created when defects are found that need to be fixed.
http://www.quickfixj.org/support/
It's a good list, and you'll learn a lot by reading questions other people have posted.

Comment by Jörg Thönnes [ 21/Jun/11 ]

Please follow Grants suggestion and post question on the list.

If you have concrete text snippets how the user manual could be improved,
please attach it to this issue. We will review it and update the manual
accordingly.

Comment by Grant Birchmeier [ 21/Jun/11 ]

I've submitted a fix to this page. It's minor, but I think it's sufficient.

Here's the diff:

  • Application application = new Application();
    + // FooApplication is your class that implements the Application interface
    + Application application = new FooApplication();
    +

Dario, what do you think?

Comment by dario marguez [ 21/Jun/11 ]

Hi, it's more clear now. Thanks for pointing me over to the mailing list.

Comment by Eric Deshayes [ 04/Aug/11 ]

This documentation improvement has been committed in rev#1047.

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