[Haskell-cafe] Why isn't pattern matching lazy by default?
Felipe Almeida Lessa
felipe.lessa at gmail.com
Wed Sep 19 06:49:06 EDT 2007
On 9/19/07, Peter Verswyvelen <bf3 at telenet.be> wrote:
> Now why isn't pattern matching lazy by default? This seems odd for a
> newbie since everything else is lazy by default.
AFAIK, pattern matches are desugared to cases, so
f (x:xs) = rhs
f  = rhs'
is equivalent to
f y = case y of
(x:xs) -> rhs
 -> rhs'
Case's have to be strict because that's how we look inside the values
in Haskell =). Of course I may be somehow mistaken here, as I am
learning Haskell, too.
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