Open alcides opened 6 months ago
Haskell supports f $ g a as a shorthand for f (g a).
f $ g a
f (g a)
The way this can be implemented is adding to the syntax the $ operator with the right precedence.
In Haskell, $ has the following type:
$
($) :: (a -> b) -> a -> b
So this depends on Polymorphism.
Haskell supports
f $ g aas a shorthand forf (g a).The way this can be implemented is adding to the syntax the $ operator with the right precedence.
In Haskell,
$has the following type:($) :: (a -> b) -> a -> bSo this depends on Polymorphism.