Control.Applicative.QQ.ADo in #haskell is
Control.Applicative.QQ.ADo
Correct Answer Show more
@brokenix the answer is in the first line of the documentation, "Pointful version of Language.Haskell.Meta.QQ.Idiom."
re: Correct Answer Show more
@antares :D i should have started the post with note to self
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re: Correct Answer
@antares :D i should have started the post with note to self