Re: [Haskell-beginners] understanding curried function calls

20 Aug 2014

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Dimitri DeFigueiredo  writes:
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Here's a simple exercise from Stephanie Weirich's class [1] that I am having
a hard time with.
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consider
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doTwice :: (a -> a) -> a -> a
doTwice f x = f (f x)
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what does this do?
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ex1 :: (a -> a) -> a -> a
ex1 = doTwice doTwice
Another way to write doTwice is:

    doTwice :: (a -> a) -> (a -> a)
    doTwice f = \x -> f (f x)

This is equivalent.  If you stare it for a while, it should answer the rest of
your questions.

John

Re: [Haskell-beginners] understanding curried function calls

John Wiegley