[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@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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