Hello,
On chapter 7 of LYH there's an example of a user-defined operator .++
infixr 5 .++
(.++) :: List a -> List a -> List a
Empty .++ ys = ys
(x :-: xs) .++ ys = x :-: (xs .++ ys)
which is used as follows
let a = 3 :-: 4 :-: 5 :-: Empty
let b = 6 :-: 7 :-: Empty
a .++ b
(:-:) 3 ((:-:) 4 ((:-:) 5 ((:-:) 6 ((:-:) 7 Empty))))
Following this the text reads:
"Notice how we pattern matched on (x :-: xs). That works because pattern matching is actually about matching constructors. We can match on :-: because it is a constructor for our own list type ..."
Source: http://learnyouahaskell.com/making-our-own-types-and-typeclasses#recursive-data-structures
Is the operator :-: a constructor? I'm confused because the definition of :-: is not prefixed by the data keyword?
- Olumide
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