Great answer Alex. Thanks! Morten
On Nov 12, 12:34 am, Alex Osborne <a...@meshy.org> wrote: > mbrodersen wrote: > > In this simple recursive expression: > > > (def t (fn [x] (if (zero? x) 0 (+ x (t (dec x)))))) > > > The fn special form is evaluated within a context where t is not yet > > bound. > > > t is only bound AFTER fn has captured its environment. > > > In other words, the closure captured by fn doesn't know anything about > > t (or maybe only that t is unbound). > > > And yet it still works if I call t: > > > (t 5) => 15 > > > My question is how Clojure ensures that t is bound to the closure > > after the closure has already been captured? > > t is a var. The var is created before the body of def is evaluated. At > that point it is unbound. The function closes over the var not the > value of the binding. That's why you can re-evaluate a function, > changing it's behaviour, without having to re-evaluate everything that > calls it. > > You'll notice that with locals (which aren't vars, they're immutable) it > doesn't work: > > (let [t (fn [x] (if (zero? x) 0 (+ x (t (dec x)))))] (t 2)) > > Also if you try to use the var in the bindings you'll get an unbound > error, not an unable to resolve symbol error: > > (def foo foo) -- You received this message because you are subscribed to the Google Groups "Clojure" group. To post to this group, send email to clojure@googlegroups.com Note that posts from new members are moderated - please be patient with your first post. To unsubscribe from this group, send email to clojure+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/clojure?hl=en
