On Wed, Jul 15, 2009 at 6:25 AM, Eugene Kirpichov
<ekirpichov@gmail.com> wrote:
Technically, the reason is not that (++) is a function, but that it is
not a constructor of the [] type.
And, not only is it not a constructor, but it also *can't* be one,
because the main characteristic of pattern matching in Haskell is that
it is (contrary to Prolog's unification) unambiguous (unambiguity of
constructors is guaranteed by the semantics of Haskell's algebraic
datatypes).
If ++ could be pattern matched, what should have been the result of
"let (x++y)=[1,2,3] in (x,y)"?
2009/7/15 minh thu <noteed@gmail.com>:
> 2009/7/15 Magicloud Magiclouds <magicloud.magiclouds@gmail.com>:
>> Hi,
>> I do not notice this before. "fun ([0, 1] ++ xs) = .." in my code
>> could not be compiled, parse error.
>
> ++ is a function; you can't pattern-match on that.
>
> Cheers,
> Thu
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--
Eugene Kirpichov
Web IR developer, market.yandex.ru
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