Type.Operators

Documentation

type ($$) f a = f a infixr 0

The $$ operator is just like $ one but with even lower precedence level. Unlike value-level $, the type-level one has precedence level of `infixr 1` in order to be used in function arguments, like `edge :: Node $ Source a -> Node $ Target a -> a`. The $$ operator has higher precedence than `->`, so the above expression would not be valid when using it.

type ($) f a = f a infixr 1

type (&) a f = f a infixl 1