[Haskell-cafe] Expressing seq

[Chris Kuklewicz]

Chad Scherrer wrote:
> I was reading on p. 29 of "A History of Haskell" (a great read, by the
> way) about the controversy of adding seq to the language. But other
> than for efficiency reasons, is there really any new primitive that
> needs to be added to support this?
> 
> As long as the compiler doesn't optimize it away, why not just do
> something like this (in ghci)?
> 
> 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?

The (sq x) function depends on x being an instance of typeclass Eq.

Imagine a new typeclass Seq that is auto-defined for all types:

class Seq a where
  seq :: a -> (b -> b)

data Foo x = Bar x | Baz | Foo y x

The instances always use a "case" to force just enough evaluation to compute the
constructor, then return id:

instance Seq Foo where
  seq a = case a of
           Bar _   -> id
           Baz     -> id
           Foo _ _ -> id

foo1,foo2 :: Foo Int
foo1 = undefined
foo2 = Bar 1

Now "(seq foo1) b" will also be undefined which "(seq foo2) b" will be "id b"
which is "b".