Hi,
Thanks for the comprehensive and considered answers. Maybe I'm missing something but defining the original function to have two definitions with a different number of args in each causes a compiler error ie. doing
mc :: (Integral a) => a -> amc x | x < 100 = x - 10 -- 1 arg
mc = mc . mc . (+ 11) -- no args
Thanks
On Sunday, 10 April 2016, 22:29, Silent Leaf wrote:
Mike: If you seek as I think you do, to write the function mc (partially) in point-free style, you must know this style implies no arguments, or at least not all arguments, mentioned, that is for example here:
mc x | x < 100 = x - 10
mc = mc . mc . (+ 11)
The second line will only be checked for pattern matching if the first one fails, so it amounts to the "otherwise" guard as here there's no pattern, so it's a bit like the pattern that always matches (mc _ = ...)
You'll remark I did write (mc =) and not (mc x =). Point free style amounts to describing a function through a composition of other functions, in an arguments-free way, here for example, (mc . mc . (+11)) being the composition of mc twice, with the "partially-applied" function (+11) == (\x -> x + 11) == (11+). This partially applied notation works for all operators by the way.
And for the record, the whitespace operator is a pure myth. First you can remove all whitespace, it still works. Second, try using the same whitespace-induced universal right-associativity with (f a b): does it amount to (f (a b))?
The reason for this right-associativity interpretation in (mc . mc (x + 11)) is because (.) itself is right associative: right-directed greediness could we say, in the vocabulary of regular expression. It's also the case of ($), and that's why we use it to counter the natural left associativity of function application:
f $ g a == f $ (g a) == ($) f (g a) == f (g a) -- (using the definition of ($) here)
instead of
f g a == (f g) a
without using ($).
The whitespace is just a meaningless character (I guess, a set of characters) used to separate juxtaposed meaningful tokens of the language when we have either (symbol,symbol) or (nonsymbol,nonsymbol), for example respectively (!! $ /= !!$) and (f g /= fg). whenever it's a nonsymbol and a symbol, whitespace is not necessary (a+, +a).
Then there's the automatic, implicit function application between two juxtaposed non-symbolic tokens. But the whitespace has never been an operator of any kind, and is totally meaningless (and optional) in (mc . mc (x + 11)).
Especially too, it's clear no whitespace survives the tokenization during the lexical phase of the (pre?) compilation, contrarily to all real operators like (+).
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