Re: [Haskell-cafe] Pattern matching does not work like this?

Err, technically, aren't functions and constructors mutually exclusive? So if something is a function, it's, by definition, not a constructor? On Wed, Jul 15, 2009 at 6:25 AM, Eugene Kirpichov 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 :

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2009/7/15 Magicloud Magiclouds :

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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 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- Eugene Kirpichov Web IR developer, market.yandex.ru _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe