Here, too, just as in the OCaml fragment, all the calls to getter and setter are working with a single mutable variable `free_var`.
+If you've got a copy of *The Seasoned Schemer*, which we recommended for the seminar, see the discussion at pp. 91-118 and 127-137.
+
If however you called `factory` twice, you'd have different `getter`/`setter` pairs, each of which had their own, independent `free_var`. In OCaml:
let factory (starting_val : int) =
in let (adder', setter') = factory 1
in ...
- Of course, in most languages you wouldn't be able to evaluate a comparison like `getter = getter'`, because in general the question whether two functions always return the same values for the same arguments is not decidable. So typically languages don't even try to answer that question. However, it would still be true that `getter` and `getter'` (and `adder` and `adder'`) were extensionally equivalent.
+ Of course, in most languages you wouldn't be able to evaluate a comparison like `getter = getter'`, because in general the question whether two functions always return the same values for the same arguments is not decidable. So typically languages don't even try to answer that question. However, it would still be true that `getter` and `getter'` (and `adder` and `adder'`) were extensionally equivalent; you just wouldn't be able to establish so.
However, they're not numerically identical, because by calling `setter 2` (but not calling `setter' 2`) we can mutate the function value `getter` (and `adder`) so that it's *no longer* qualitatively indiscernible from `getter'` (or `adder'`).