Re: [Haskell-cafe] How to understand the 'forall' ?

seems a bit understanding, i still need to think it for a while thanks! jkff wrote:

2009/9/2 zaxis :

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Isnot it clear without the 'forall' ? data Branch tok st a = Branch (PermParser tok st (b -> a)) (GenParser tok st b)

thanks!

The situation is not so simple. Consider, for example, a procedure that takes as input a *generic* sorting algorithm:

sortThem :: (forall a. Ord a => [a] -> [a]) -> [Int] -> [String] -> ([Int], [String]) sortThem sortAlgo ints strings = (sortAlgo ints, sortAlgo strings)

Here you can't omit the 'forall' because if you do, then inside the body of the sortThem function the 'a' type variable has a fixed value and can't be both Int and String!

This is a somewhat esoteric example, but one could consider, for instance, a procedure that takes as input some trees of different types and a generic tree traversal algorithm. Well, have a look at the haskellwiki page, there's a pile of examples :)

P.S. I tried to write up the difference between datatype and function declarations in this respect, but my explanations turned into a mess, so I erased them in the hope that someone will explain it better than me.

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jkff wrote:

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This means that for any type 'b' you can construct a value of type 'Branch tok st a' by passing to Branch an argument of type '(PermParser tok st (b -> a))' and 'GenParser tok st b'. This also means that when you're given a value of type Branch tok st a, you don't know what that 'b' type was; the only thing you know is that the 'b' in 'b -> a' in the first argument of Branch is the same as the 'b' in 'GenParser tok st b'.

See also: the haskellwiki page on existential types.

2009/9/2 zaxis :

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data Branch tok st a = forall b. Branch (PermParser tok st (b -> a)) (GenParser tok st b)

please shed a light on me, thanks! -- View this message in context: http://www.nabble.com/How-to-understand-the-%27forall%27---tp25250783p252507... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.

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