# seq/parametricity properties/free theorems and a proposal/question

Fri Mar 18 15:54:59 CET 2011

```The haskell (2010) report says:

"The function seq is defined by the equations:

seq _|_ b  =  _|_
seq a b  =  b, if a /= _|_

seq is usually introduced to improve performance by avoiding unneeded
laziness. Strict datatypes (see Section 4.2.1) are defined in terms of the \$!
operator. However, the provision of seq has important semantic consequences,
because it is available at every type. As a consequence, _|_ is not the same
as \x -> _|_ since seq can be used to distinguish them. For the same reason,
the existence of seq weakens Haskell’s parametricity properties."

As has been noted in the literature (and presumably the mailing lists), this

map f . map g = map (f . g)

should be derivable from the type of map "(a -> b) -> [a] -> [b]" (i.e., as
map works on all a's and b's, it can't actually do anything other than move
the a's around, wrap them with the given function to get the b's, or pass them
along to other functions which should be similarly constrained).

However, this last bit about passing them to similarly constrained functions
is not true thanks seq.  There is no guarantee map, or some other polymorphic
function it invokes has not used seq to look inside the universally quantified
types and potentially expose _|_.  For example, consider

map' f [] = []
map' f (x:xs) = x `seq` f x : map f xs

Now "map' f . map' g = map' (f . g)" is not true for all arguments.

map' (const 0) . map' (const undefined :: Int -> Int) \$ [1..10] = [undefined]
map' (const 0 . undefined :: Int -> Int) \$ [1..10] = [0,0,0,0,0,0,0,0,0,0]

My proposal (although I'm sure someone must have brought this up before, so
it's more of a question) is why not take the magic away from seq by making it
a class function like deepSeq

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

It is easily implemented in haskell by simply forces the constructor for the
underlying data type (something also easily derivable by the compiler)

instance Seq Int where
seq' 0 y = y
seq' _ y = y

Now, unless I'm missing something, we have restore the strength of Haskell's
parametricity properties

map :: (a -> b) -> [a] -> [b]
map f [] = []
map f (x:xs) = f x : map f xs

map' :: Seq a => (a -> b) -> [a] -> [b]
map' f [] = []
map' f (x:xs) = x `seq'` f x : map f xs

as the full range of operations that can be performed on universally quantified
types (in addition to moving them around) is reflected in any explicitly and
implicitly (through the dictionaries) passed functions over those types.

Thanks!  -Tyson
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