On Dec 24, 2011 12:53 PM, "Benjamin Edwards" wrote:

Because patterns are matched in order and the first one match is taken.

This is very expressive. For example, you can write f 0 = 0 f x = 1 / x

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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