teaching programming to kids - Re: Looking for 8080/Z80 BASIC

In some respects. However, each closure contains its own distinct set of values; most languages' "static local variables" don't work that way. As a really simple example, consider (to use Lisp as the language, since that's what I'm most familiar with practical closures in; I'm using my own Lisp here, which is pretty close to Zetalisp in the relevant respects): (defun make-adder (n) #'(lambda (x) (+ x n))) Then (setq a4 (make-adder 4)) (setq a10 (make-adder 10)) a4 and a10 are then closures. Each one has (lambda (x) (+ x n)) as its code blob; its binding list includes a binding of n to some value - 4, in the case of a4, or 10, in the case of a10. (funcall a4 7) -> 11 (funcall a10 7) -> 17 Whereas, in C, which doesn't have closures but does have static local variables, static int n; static int adder(int x) { return(x+n); } static int (*make_adder(int val))(int) { n = val; return(&adder); } the function returned by make_adder will always add the last value passed to make_adder, regardless of which call's return value is called. That is, after int (*a4)(int); int (*a10)(int); a4 = make_adder(4); a10 = make_adder(10); Then (*a4)(7) gives 17, not 11 - though if called between the two make_adder calls, it returns 11. This is because the returned pointers are _not_ closures; the binding of the name n in adder to the storage location it refers to is fixed, and all those references refer to the same location. (Actually, I've seen it said that there's an effort afoot to get closures into C, along with anonymous code blocks and various other things. I think that is spectacularly misguided; if they want Lisp, they know where to find it.)