Ah, thanks... However, I think I have just become more confused as to what makes a closure a closure. The defining principle seems to go along the lines some kind of context that is enclosed within the closure object. It was argued elsewhere [1] that a higher order function is not necessarily a closure. So, coming back to the Wiki example of f x = (\y -> x + y)
The free variable x in the lambda term refers to a variable bound outside of the actual term. However, I don't quite see how this can be seen as enclosing some kind of context... it's just a function returning a function, isn't it? This seems to be underlined by the fact that it can be rewritten in a pointfree fashion where there really is no binding whatsoever at the time of definition. If someone could point out a more specific example for showing where the difference between closures and higher-order functions could be drawn I'd be most grateful... [1] http://notes-on-haskell.blogspot.com/2007/02/whats-wrong-with-for-loop.html#comment-859881452769224042 On 9/3/07, Derek Elkins <[EMAIL PROTECTED]> wrote: > On Mon, 2007-09-03 at 19:47 +0200, Lars Oppermann wrote: > > Dear all, > > > > In the Haskell Wiki at http://www.haskell.org/haskellwiki/Closure > > there is an example for a function returning a closure given as > > f x = (\y -> x + y) > > > > Another way to achieve the same effect would be to write > > f' x = (+) x > > > > which to me as a beginner looks somewhat like pointfree style. > > > > Would f' be considered to return a closure > > Going backwards: > > Yes. > > The pointfree "solution" in this case would simply be f' = (+) > > Note that f x = \y -> x + y is written that way for emphasis and is > identical to f x y = x + y > > _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
