Re: [Haskell-cafe] Expressing seq

Chad Scherrer wrote:

Prelude> let sq x y = if x == x then y else y Prelude> 1 `sq` 2 2 Prelude> (length [1..]) `sq` 2 Interrupted.

There must be a subtlety I'm missing, right?

Two, at least: First, your sq has a different type, as it requires an Eq instance: Prelude> :t sq sq :: (Eq a) => a -> t -> t Prelude> :t seq seq :: a -> b -> b Secondly, your sq is more akin to a deepSeq in that it forces all of its value instead of just evaluating to weak head normal form. Prelude> [undefined] `seq` 1 1 Prelude> [undefined] `sq` 1 *** Exception: Prelude.undefined You could implement seq explicitely for many types, for example, seqList [] x = x seqList (_:_) x = x but not for function types. HTH, Bertram