
Are you saying that GHC doesn't have a haskell parser?
On Mon, Jun 8, 2015 at 12:06 PM, Rein Henrichs
The point is that there is no parsing happening here. Just evaluation.
On Mon, Jun 8, 2015 at 3:20 AM Mike Meyer
wrote: On Jun 8, 2015 05:01, "Martin Vlk"
wrote: Mike Meyer:
s mp $ n - 1 parses as s mp (n - 1) s mp n - 1 parses as (s mp n) - 1
Hi, I think it is misleading to say that $ "parses" to something.
Which is why I didn't say such a thing. The two sentences show how the expressions he was having trouble with are parsed.
($) is an infix operator with type:
($) :: (a -> b) -> a -> b -- Defined in ‘GHC.Base’ infixr 0 $
So what it does is it takes a function on the left hand side and applies it to the second argument on the right hand side. The trick is it has the lowest possible precedence value (0) so everything on the right hand side will be evaluated before applying the function on the left.
Which explains why the expression containing a $ parses as it does, which is also useful information. _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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