Comments on: Coding Dojo #13: TDD as if you meant it

By: Coding Dojo #14: Rock, Scissors, Paper - TDD as if you meant it at Mark Needham

Coding Dojo #14: Rock, Scissors, Paper - TDD as if you meant it at Mark Needham — Thu, 14 May 2009 21:40:59 +0000

[...] attempt a fortnight ago was around implementing a Flash Message interceptor, to hook into the Spring framework but this [...]

By: Keith Braithwaite

Keith Braithwaite — Thu, 14 May 2009 15:42:26 +0000

@Mark,
Good, that's the feeling I look for in the exercise.

By: Keith Braithwaite

Keith Braithwaite — Thu, 14 May 2009 15:41:51 +0000

@Bill,
The compiler rejects the call to foo.bar() because bar is a method on an un-named sub-class of Foo, but foo is declared as a pointer to Foo so the method is not visible via that pointer.

If you are following the intent of the exercise, you will not create a method bar like this, you will write some code in the test method that does the work and then (if it turns out to be desirable) factor that code out into a method–and perhaps not on Foo.

Hope that helps.

By: Mark Needham

Mark Needham — Fri, 01 May 2009 14:43:46 +0000

@Keith – I felt pretty good about the code, it felt much more like it evolved doing it this way.

I discussed this with a few guys in the Dojo and we thought that the FlashMessageStore in this case became more obvious and that if we'd have coded the normal way we might not have seen that abstraction. Certainly an exercise to try again

By: Bill Comer

Bill Comer — Fri, 01 May 2009 14:36:17 +0000

Mark,

I thought I would give this a go in Java and failed, but I am unsure quite why. Perhaps you can enlighten me.

I have a class Foo with some existing methods on it.
I wish to add a new method bar() to do some work.
Without alterring Foo I go to my test class and add a new test:

public void testBar() throws Exception {
Foo foo = new Foo() {
public void bar(){
// do some Baaing..
}
}
foo.bar();
}

The compiler happily accepts the method bar() but fails to compile with a method undefined for the line:
foo.bar();

If I create an interface with bar() in or a stub in Foo then all is happy but this appears to miss the point.

By: Keith Braithwaite

Keith Braithwaite — Wed, 29 Apr 2009 20:29:27 +0000

Mark,
Interesting experience, thanks for writing it up. Now that you've been through the exercise on a realistic problem, how do you feel about the resulting code? (Note: "how do you feel?" not "what do you think?)