Lazy suspended tail in sml

Question

I was going through some notes and I realized something is amiss.

When emulating lazy computation (without open Lazy;) one can do the following for a stream of ones.

datatype 'a susp = Susp of (unit -> 'a)

datatype 'a stream' = Cons of 'a * ('a stream') susp
type 'a stream = ('a stream') susp

fun delay (f ) = Susp(f);

fun force  (Susp(f)) = f ();

val rec ones' = fn () => Cons(1, delay(ones'));

val ones = delay(ones')

fun ltail(Susp(s)) = ltail'(force s)   
and ltail' (Cons(x,s)) = s

But for getting a suspended tail the types do not match up.

operator domain: 'Z susp   
operand:         unit -> 'Y

What will need to change for the proper types for ltail ? I know what happens with a tail not suspended. I just want to figure out what the notes were saying for the suspended version.

Interesting language. Never heard of it. Since when does NJ get their own language and Texas doesn't? :P — Jared Farrish, Mar 15 '11 at 00:30

score 2 · Accepted Answer · answered Mar 15 '11 at 01:30

2

fun ltail(Susp(s)) = ltail'(force s)

The problem here is that force takes a value of type susp, but you call it with a value of type () -> 'a. I.e. you take the function out of the susp value and then call force on the function instead of the susp value. You should just do:

fun ltail s = ltail' (force s)

answered Mar 15 '11 at 01:30

sepp2k

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Yeah I tried that when I sat down with the notes , though I wasn't sure why it was with Susp(s) in the notes. Oh well ... I will assume it was a typo and life moves on. – phwd Mar 15 '11 at 02:15

Lazy suspended tail in sml

1 Answers1