Get the root of a Rose tree in Haskell

Question

Recently I started to learn about Haskell, and am struggling with the following exercise:

Write functions root :: Rose a -> a and children :: Rose a -> [Rose a]
that return the value stored at the root of a rose tree, respectively the children of the
root of a rose tree.

They gave me the following basic code to start:

data Rose a = a :> [Rose a]
    deriving (Eq, Show)

I don't know what (:>) means. I tried to understand it by typing in the ghci

Input: :t (:>)
Output: a -> [Rose a] -> Rose a

So I think it means that you have an a value, which will be used to lookup Rose a out of a list and returns Rose a, but I'm still confused what to do next.

If I look at the signature of root :: Rose a -> a, the function would look like this:

root (Rose a) = a 

And the function of children :: Rose a -> [Rose a]:

children (Rose a) = (Rose a):[]

This is not correct, and I don't how to make it correct.

Answer 1

The declaration

data Rose a = a :> [Rose a]

is basically equivalent to

data Rose a = Node a [Rose a]

In other words, Node is a data structure containing a datum and a list of child nodes. But in the definition above, rather than calling it Node, it's called :>. It's just a made-up name; Haskell allows you to create user-defined operators like this.

If the name Node was used, you could write

root (Node datum children) = datum

Or, more briefly,

root (Node a rs) = a

Since the name given is actually :>, you'd have to write it as

root (a :> rs) = a

In particular, you seem to be trying to use Rose, but that's the type constructor, not the value constructor. Similarly, you seem to be trying to use the ":" operator, but that's for lists, not rose trees.

Hopefully that clears up some things for you.