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#...
On 9/3/07, Derek Elkins
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