Get the root of a rose tree in Haskell
I recently started learning 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:
data Rose a = a:> [Rose a]
deriving (Eq, Show)
I don't know what (:>) means. I tried to figure it out by typing ghci
Input: :t (:>)
Output: a -> [Rose a] -> Rose a
So, I guess this means you have a value a
to be used to search Rose a
from the list and return Rose a
, but I'm still confused as to what to do next.
If I look at the signature root :: Rose a -> a
, the function looks like this:
root (Rose a) = a
And the function children :: Rose a -> [Rose a]
:
children (Rose a) = (Rose a):[]
This is not correct and I don't understand how to do it.
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Ad
data Rose a = a :> [Rose a]
basically equivalent to
data Rose a = Node a [Rose a]
In other words, it Node
is a data structure containing a database and a list of child nodes. But in the above definition, instead of naming it Node
, it was called :>
. It's just a fictitious name; Haskell allows you to create custom operators like this.
If a name is used Node
, you can write
root (Node datum children) = datum
Or, more briefly,
root (Node a rs) = a
Since the name is actually given :>
, you need to write it as
root (a :> rs) = a
In particular, you are trying to use Rose
, but it is a type constructor, not a value constructor. Similarly, you are trying to use the " :
" operator , but for lists, not rose trees.
Hope this clears things up for you.
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