>From my moving-target understanding of closures, I'm guessing this has a
closure
let succ = add 1
add x = xadder
where xadder y = x + y
in succ 3
Now, can someone explain in words how the closure system is at work here? I
see that the add 1 is being "baked in". Do we say this is a closure on add 1?
(Closure wording is confusing.) The enclosing scope of add contains 1;
however, xadder is what succ ultimately becomes, which has 1 in its scope
in that when xadder is defined, the x argument will be the 1 of add 1. Or
am I missing something?
Researching closures vis-a-vis Haskell, I've seen the argument that,
- no, Haskell doesn't have closures,
- yes, Haskell is lazy, *everything* can be seen as a closure, as even a
value can be seen as a function without argument waiting to be evaluated
(and so capturing its environment until it gets evaluated).
This was taken from an older Haskell textbook (*Introduction to Functional
Programming Systems Using Haskell *by Davie).
LB
