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

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

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 :

...

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

-- View this message in context: http://www.nabble.com/How-to-understand-the-%27forall%27---tp25250783p252517... Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.