newtype/datatype (was efficiency)
Ken Shan
[email protected]
Wed, 16 Jan 2002 23:25:43 -0500
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A while ago, to help myself understand newtype, data, and strictness,
I tried to write down how Haskell types correspond to lifted domains.
Here is a cleaned-up version of my attempt. I am not sure that what
follows is entirely correct -- please point out any errors. I would
also appreciate comments or pointers to where this stuff is better
described. In particular, I have not considered recursive types.
Every type "t" in Haskell can be denotationally modeled by a complete
partial order (CPO), which we will notate as "[t]". In fact, every
type in standard Haskell corresponds to a -pointed- CPO, meaning a CPO
with a least element "bottom". GHC, in addition, supports -unboxed-
types, which correspond to CPOs in general.
Given a CPO "D", we can -lift- it to a pointed CPO "lift D" by adding
a bottom.
Given two (not necessarily pointed) CPOs "D" and "E", we can take
their -cartesian product- "D x E", which is like the set-theoretic
product.
Given two pointed CPOs "D" and "E", we can take their -smash product-
"D * E", which is like the set-theoretic/cartesian product, but which
identifies all pairs of the form "(x,bottom)" or "(bottom,y)" into one
single new bottom element. The identity of "*" is the two-element
pointed CPO "1", which consists of a bottom and a top.
Given two pointed CPOs "D" and "E", we can take their -coalesced sum-
"D + E", which is like the set-theoretic sum, but which identifies the
bottom from "D" and the bottom from "E" into one single new bottom
element. The identity of "+" is the one-element pointed CPO "0",
which consists only of a bottom.
Given any two (not necessarily pointed) CPOs, we have the property
lift D * lift E =3D lift (D x E).
In Haskell, if you write
newtype t' =3D T t -- ,
then "[t']" is exactly isomorphic to "[t]". Pattern matching on the
"T" is a no-op (i.e., always succeeds).
Now turn to a data declaration
data t' =3D T1 ... | T2 ... | ... -- .
Each branch in the declaration specifies a pointed CPO by naming zero
or more data types, each of which can be strict or non-strict. We
lift the non-strict ones, then take the smash product of everything.
The result is the pointed CPO denoted by the branch. The pointed CPO
"[t']" denoted by the combined data type "t'" is the coalesced sum of
the pointed CPOs denoted by all the branches.
For example, if we write
data t' =3D T1 !t11 !t12 | T2 t2 | T3 -- ,
then
[t'] =3D [t11] * [t12] + lift [t2] + 1 -- .
Pattern-matching on any data-declared type such as "t'" above is
always strict, in the sense that if the value being matched against is
bottom, then the whole expression denotes bottom. For example, given
how we defined "t'" above, the expression
case x' of T1 _ _ -> y
T2 _ -> y
T3 -> y
is entirely equivalent to
x' `seq` y -- .
The only reason to have `seq` is because primitive types such as
Int cannot be pattern-matched against. (Hence people call "seq"
"polymorphic case".)
Unboxed types are non-pointed CPOs, so it doesn't make sense to say
data t' =3D T !t
if "t" is an unboxed type, but it does make sense to say
data t' =3D T t
because "[t]" can be lifted to a pointed CPO. You ought to be able to
say
newtype t' =3D T t
to define a new unboxed type "t'" from an existing unboxed type "t",
but I faintly remember this being not implemented or allowed or
something.
Tuple types are special notation for what could be otherwise defined
using "data" declarations as follows:
data () =3D ()
data (,) t1 =3D (,) t1
data (,,) t1 t2 =3D (,,) t1 t2
data (,,,) t1 t2 t3 =3D (,,,) t1 t2 t3
-- ...
So we have
[( )] =3D 1
[(t1 )] =3D lift [t1]
[(t1,t2 )] =3D lift [t1] * lift [t2]
=3D lift ([t1] x [t2])
[(t1,t2,t3)] =3D lift [t1] * lift [t2] * lift [t3]
=3D lift ([t1] x [t2] x [t3])
...
If "[t1]" has 3 elements including its bottom, and "[t2]" has 5
elements including its bottom, then "[(t1,t2)]" has 3 * 5 + 1 =3D 16
elements including its bottom.
Given the definitions above, let us consider some corner and not so
corner declarations: (When I write that foo denotes bar, I mean that
foo denotes something isomorphic to bar.)
data T1 -- denotes "0", the identity for "+"
-- (but this is not standard Haskell)
data T2 =3D T2 -- denotes "1", the identity for "*"
data T3 =3D T3 () -- denotes "lift 1", because "()" denotes "1"
data T4 =3D T4 !() -- denotes "1", because "()" denotes "1"
newtype T5 =3D T5 () -- denotes "1", because "()" denotes "1"
data T6 =3D T6 !T1 Int -- denotes "0 * lift [Int]", which is just "0"
Note that although both T4 and T5 denote "1", pattern-matching on them
is different:
case T4 undefined of T4 _ -> y =3D=3D=3D undefined
case T5 undefined of T5 _ -> y =3D=3D=3D y
Whew.
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