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