The operator ($) is often considered an application operator of a
lower precedence. Modulo precedence, there seem to be no difference
between ($) and `the white space', and so one can quickly get used to
treat these operators as being semantically the same. However, they
are not the same in all circumstances. I'd like to observe an
important case where replacing the application with ($) in a
fully-parenthesized expression can lead to a type error.
> {-# OPTIONS -fglasgow-exts #-}
>
> module Foo where
>
> data WR = WR (Int -> Int)
> data W = W (forall a. a->a)
>
> t1 = WR id
> t2 = W id
We can also write
> t1' = WR $ id
However, if we try
> t2' = W $ id
we get an error:
/tmp/t1.hs:13:
Inferred type is less polymorphic than expected
Quantified type variable `a' escapes
Expected type: (a -> a) -> b
Inferred type: (forall a1. a1 -> a1) -> W
In the first argument of `($)', namely `W'
In the definition of `t2'': t2' = W $ id
Incidentally, Hugs -98 gives a quite bizarre error message
ERROR "/tmp/t1.hs":13 - Use of W requires at least 1 argument
It didn't complain about "WR $ id"...
The reasons for that behavior are obvious: the compiler cannot
generalize to higher-ranked types when instantiating the type of
($). It makes a difference that the application is a built-in
construction, whereas ($) is just a regular function.
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