[Haskell-beginners] my own last function

[David McBride]

[a] when used as an argument doesn't mean what you think it does.  It
means a list with only one element a.  So your function will only
match one single case, a list such as [3], or a string with one
character "a".

Here are all the things you could match on:

xs -> just a variable which will contain the whole list
[] -> an empty list
[a] -> a list with one element in it, which you can reference via a
[a,b,c] -> a list with exactly three elements in it, which can all be referenced
(x:xs) -> a list with at least one element, of which x is that
element, and xs is the rest of the list
(x:[]) -> a list with one element, followed by nothing else.
Essentially the same as [x].
(x:y:xs) -> a list with two elements, xs has the rest of the list.
Remember that xs could be empty.

So think about the problem.  If someone passes in a list, it either
has one element, or it has more than one and we want to get to the
last one
last :: [x] -> x
last [x] = x

Well that deals with the easy case.  One element, one answer.  But if
there is more than one element, then what?
last (_:xs) = last xs

Just discard the first one and run the function again on what remains.


On Sat, Feb 18, 2012 at 1:18 PM, bahadýr altan <doaltan at yahoo.co.uk> wrote:
> Hello everyone, I have just started studying and started trying to solve 99
> questions in haskell.org. However I got stuck even in the first question.
> The first question wants me to write my own last function and I wrote this
> :
>
> q1mylast :: [x] -> x
> q1mylast [a] = if length [a] <=1 then  a
> else  q1mylast (tail [a])
>
> This program doesn't work and it gives this error :
>
> Program error: pattern match failure: q1mylast "doaltan"
>
> I'll be very glad if you could tell me what's wrong with the code and how
> can I fix it..
> Thanks in advance :)
>
>
>
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