Re: [Haskell-beginners] Right-associating infix application operators
Hi Tom, There are no additional benifits to the '$' function - you pretty much have it. Wait - one more. It can be used as a 'section' like any other binary operator, so ($ 4) :: (Int -> a) -> a. But that doesn't come up a whole lot. This is much like how (* 5) 6 == 30. Antoine On Jul 6, 2010 6:00 AM, "Tom Hobbs" <tvhobbs@googlemail.com> wrote: In people's responses to my serialization questions, I've seen them using $. I didn't know what it was so I've looked it up. Can someone please confirm my understanding of what it does, please? According to http://en.wikibooks.org/wiki/Haskell/Practical_monads, after the second code sample in the "Return Values" section, it seems to suggest that $ is only used to avoid using so many brackets. Which seems to make sense, but looking at it's definition in Prelude I really can't see why it's useful. Yitz gave me the code; fmap (runGet $ readNames n) $ L.hGetContents h So can I rewrite this without the $ like this? fmap (runGet (readNames n)) (L.hGetContents h) Is there any additional benefit to using $ than just not having to write as many brackets? Thanks, Tom _______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners
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Antoine Latter