How would we take the last value of a list using colon notation?

Question

Say we have a list in the parameter and pattern match on its head and tail separately so we use x:xs. This means that the inputs are split into a value x and a list xs.

So whatever is before the colon is treated as a single value and whatever is last in the colon notation will be the rest of the list with the single value x removed.

But what if I wanted to separate the last value from the rest of the values? It seems that this notation cant be used.

Update: i think i know how i would be able to find the last number using recursion, but is the true that you can't simply get it through some kind of pattern matching?

Cactus Cactus · Accepted Answer · 2016-03-30T02:10:36

The Haskell datatype [a] represents lists as a singly-linked cons list: the pattern match x:xs is not merely notation, but that's really how it is stored in memory, as a cons of the head and the rest.

As you mentioned in your update, you can write a function which goes through a list and splits it into an init part and a last element:

unsnoc :: [a] -> Maybe ([a], a)

You can use a view pattern to use it in pattern matching:

f (unsnoc -> Just (xs, x)) = ...
f [] = ...

And you can even wrap it into a pattern synonym:

pattern xs :> x <- (unsnoc -> Just (xs, x))

(or the bidirectional version

pattern xs :> x <- (unsnoc -> Just (xs, x)) where
  xs :> x = xs ++ [x]

)

allowing you to write

f (xs :> x) = ...
f [] = ...

Note however that all these approaches still require traversing the spine of the whole list, since they are all ultimately just calling your unsnoc function.

How would we take the last value of a list using colon notation?

2 Answers