[Haskell-cafe] A probably-stupid question about a Prelude
phil at kantaka.co.uk
Fri Jun 22 11:43:45 EDT 2007
On Fri, Jun 22, 2007 at 11:31:17PM +0800, Michael T. Richter wrote:
> 1. Using foldr means I'll be traversing the whole list no matter what.
> This implies (perhaps for a good reason) that it can only work on a
> finite list.
foldr is lazy.
> Please tell me I'm wrong and that I'm missing something?
You are wrong and you're missing something :)
any ((==) 2) [1,2,3]
any ((==) 2) [1..]
any ((==) 0) [1..] will go _|_ of course.
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