[Haskell-beginners] Closure clear-up?

[Galaxy Being]

>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
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