[Haskell-cafe] Conditional code for Haddock

Roel van Dijk vandijk.roel at gmail.com
Wed Jan 20 07:40:20 EST 2010

> 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

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