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 <z_a...@163.com>: >> >> 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---tp25250783p25250783.html >> Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. >> >> _______________________________________________ >> 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 > > -- View this message in context: http://www.nabble.com/How-to-understand-the-%27forall%27---tp25250783p25251783.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
