Defunctionalization is a method of converting a functional program into one that involves no higher order functions nor lambda expressions. It is a whole program transformation.
The idea is it to consider a type $Defun(A,B)$ for every pair of types $A,B$ as a proxy for the true function type $A\to B$. This type $Defun(A,B)$ is a coproduct
over all lambda abstractions $\lambda y.\,t$ that appear syntactically in the program. Notice that lambda abstractions may have free variables $x_1:C_1,\dots,x_n:C_n$, because they may appear deep inside the program. An inhabitant of the type $Defun(A,B)$ is sometimes called a closure, because it is a syntactic expression together with an environment: a valuation for its free variables.
The type $Defun(A,B)$ has a canonical evaluation function $eval:Defun(A,B)\times A\to B$ roughly given by
and so it behaves like a function space.
Since $A$, $B$ and the $C_i$’s can also be defunctionalized, this is a method to transform a whole functional program into one with no lambda expressions nor any true function types.
The coproduct type $Defun(A,B)$ is typically not a true function space in the sense of cartesian closed categories for two reasons.
it does not admit arbitrary lambda abstractions – just the lambda abstractions that happen to be included in the coproduct.
it does not equate lambda abstractions that are semantically equal but syntactically different.
(Thus it violates both existence and uniqueness in the universal property.)
Recall that a cartesian internal hom $B^A$, if it exists, is a representation of $Hom(-\times A,B)$. As such, in many concrete situations, we can deduce its existence from an adjoint functor theorem. Recall that the solution set condition in this instance requires a set $I$ and an $I$-indexed family of objects $C_i$ together with morphisms
such that every morphism $f:C\times A\to B$ factors as
for some $i\in I$ and $g\colon C\to C_i$. Then, as is usually observed, the coproduct
is a weak internal hom, from which the internal hom is recovered by taking a coequalizer. This coproduct is strongly reminiscent of the coproduct type $Defun(A,B)$ used in defunctionalization.
Consider the factorial function in continuation-passing style, written in Haskell:
factcps :: Integer -> (Integer -> Integer) -> Integer
factcps x k = if x == 0 then k 1 else factcps (x-1) (\y -> k (x * y))
fact :: int -> int
fact x = factcps x (\x -> x)
The idea is that factcps x k
calculates the factorial of x
but does not return the result, instead, it calls the “continuation” k
with the result. Since fact 5
calls factcps 5
with the identity function, it will eventually return 120
.
If this is the whole program, then there are only two lambda abstractions we need to deal with, both of type Integer -> Integer
. The first one \y -> k (x * y)
has free variables x
and k
; we’ll call it Multiply
. The second \x -> x
has no free variables; we’ll call it Identity
.
To defunctionalize this we define consider a type DefunIntInt
, which is $Defun(Integer,Integer)$ above:
data DefunIntInt = Multiply Integer DefunIntInt | Identity
eval :: DefunIntInt -> Integer -> Integer
eval (Multiply x k) y = eval k (x * y)
eval (Identity) x = x
And now we can write the program without lambda abstractions:
factcps :: Integer -> DefunIntInt -> Integer
factcps x k = if x == 0 then eval k 1 else factcps (x-1) (Multiply x k)
fact :: Integer -> Integer
fact x = factcps x Identity
In fact, in this example, the type DefunIntInt
is isomorphic to the type of integer lists, and squinting slightly it can be seen as a call stack.
Defunctionalization was probably introduced by John Reynolds in
There and elsewhere a common theme is to first convert programs to continuation-passing style, which makes the control stack explicit, and then to defunctionalize, which makes the control stack look like a stack; the net result is a kind-of abstract machine.
Last revised on July 8, 2019 at 13:26:38. See the history of this page for a list of all contributions to it.