Inferring recursive expressions using Hindley Milner & constraints

Question

I am trying to infer the type of the following expression:

let rec fix f = f (fix f)

which should be given the type (a -> a) -> a

After using the bottom up algorithm (described in generalizing hindley-milner type inference algorithms) with the added rule below:

           a1, c1 |-BU e1 : t1     B = fresh var
---------------------------------------------------------
a1\x, c1 U {t' == B | x : t' in A} |-BU let rec x = e1 : t

i am left with the following type: t1 -> t2

and the following constraints:

t0 = fix
t1 = f
t2 = f (fix f)
t3 = f
t4 = fix f
t5 = fix
t6 = f

t3 = t1
t3 = t4 -> t2
t5 = t0
t5 = t6 -> t4
t6 = t1

I cant see how these constraints can be solved such that i am left with the type (a -> a) -> a. I hope it is obvious for someone to see were i am going wrong.

full source code here

Please note that let rec should be simply let, otherwise you're just defining a function rec :: ((a -> b) -> a) -> (a -> b) -> b. — Ptharien's Flame
@Ptharien'sFlame sorry code example wasnt in haskell but an ML style language. — user181351
Okay, it's just that you had the question tagged with "haskell" so I assumed that was the language you were working with. Sorry! — Ptharien's Flame

Sjoerd Visscher Sjoerd Visscher · Accepted Answer · 2012-05-13T21:40:33

Shouldn't there be a t7 for the first fix f? These give the constraints:

t7 = t2
t0 = t1 -> t7

From that you should be able to deduce that t4 = t2 and then t0 = (t2 -> t2) -> t2.

Inferring recursive expressions using Hindley Milner & constraints

1 Answers