Re: [Haskell-beginners] Prelude Implementation of Last overlapping patterns?

Because patterns are matched in order and the first one match is taken. On 24 Dec 2011 17:49, "Mark Stoehr" wrote:

The standard Prelude implementation of 'last' is as follows \begin{code} last :: [a] -> a last [x] = x last (_:xs) = last xs last [] = error "Prelude.last: empty list" \end{code}

Isn't [x] equivalent to (x:[]) hence wouldn't [1] match both [x] and (_:xs)? If that's the case then we would have last [1] == 1 and last [1] == last []

but that doesn't happen. Why?

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