[Haskell-cafe] Conditional code for Haddock

[Roel van Dijk]

> Doesn't haddock define __HADDOCK__ by itself?

That appears to be a common misconception. The discussion on this
ticket [1] indicates that haddock does *not* define __HADDOCK__. So I
can not rely on it being defined. Therefore I would like to define it
myself, but only in the case that haddock is being run on my code.


1 - http://trac.haskell.org/haddock/ticket/76