I'm not sure what you mean by "abstract function". It's a function like any other function. But otherwise you're right. :)
-- Lennart On 9/22/07, PR Stanley <[EMAIL PROTECTED]> wrote: > > Ah, I understand now. Let me get this right: > The resulting (a -> c) is a kind of abstract function formed by the > pairing of <a -> b) and (b -> c), hence the lambda \x -> g (f x), correct? > Thanks, Paul > > At 05:11 22/09/2007, you wrote: > >Hello, > > > >It's probably easiest to think of composition as a function which > >takes two arguments (both functions), (g :: b -> c) and (f :: a -> b), > >and returns a new function of type a -> c. We could write this > >explicitly as > > > > composition :: (b -> c, a -> b) -> a -> c > > composition (g,f) = \x -> g (f x) > > > >then (.) is the currying of composition: > > > > (.) = curry composition > > > >or > > > > (.) g f = \x -> g (f x) > > > >-Jeff > > > > > >On 9/21/07, PR Stanley <[EMAIL PROTECTED]> wrote: > > > Hi > > > (.) :: (b -> c) -> (a -> b) -> (a -> c) > > > While I understand the purpose and the semantics of the (.) operator > > > I'm not sure about the above definition. > > > Is the definition interpreted sequentially - (.) is a fun taht takes > > > a fun of type (b -> c) and returns another fun of type (a -> b) etc? > > > Any ideas? > > > Thanks, Paul > > > > > > _______________________________________________ > > > Haskell-Cafe mailing list > > > [email protected] > > > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > > > _______________________________________________ > Haskell-Cafe mailing list > [email protected] > http://www.haskell.org/mailman/listinfo/haskell-cafe >
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