Thanks, i think i understand that (lambda (e) (e x)) is a function, and that evaluating it produces a new function (the one we use in the recursive call - let's call it rf1).
Is your meaning that rf1 protects the self-replicator embedded in the 'else' clause from triggering, and so protects us from ongoing re-production? When we get to it, a fresh call to (lambda (e) (e x)) --> rf2, which again protects the replicator (the mental image i have is of a popcorn kernel awaiting the heat). Thanks for your patience & sorry if i am being a little dense - this is all very new to me. best regards mj On 11/02/2014, at 3:23 AM, Hendrik Boom <hend...@topoi.pooq.com> wrote: > On Mon, Feb 10, 2014 at 09:52:18PM +0000, Matthew Johnson wrote: >> Thank-you all for your help with my Y-combinator question. >> >> I've now got myself a copy of *The Little Schemer*, and think that i'm >> getting there. >> >> Some googling on the lambda calculus and fixed point combinators introduced >> me the formalism: >> >> ((lambda (x) (x x)) (lambda (x) (x x))) >> >> The general setting helped me see what was going on with (make-list >> make-list) etc ... in an abstract sense (i've never studied CS) however i >> do remain a little puzzled by Matthias' second hint: >> >> "when you have an expression e and you know it will evaluate to a function >> if the evaluation terminates but it may not terminate, you can write >> (lambda (x) (e x)) to make sure that it is a function NOW. That way you can >> pass this quasi-e to another function and it can make the decision to >> evaluate it or not. If you had written the derivation in #lang lazy as >> opposed to #lang racket, you would not need this trick. In a lazy language >> e is passed as an argument without being evaluated to start with." >> >> Why does (lambda (x) (e x)) make it evaluate once and stop? I mean how come >> when the evaluator gets to the call it doesn't get stuck in a loop? > > Because when you evaluate (lambda (x) (e x)) you get a function. If > that function is ever called, it will take an argument x, evaluate the > expression e, and then call the value of e passing it x as argument. > If evaluating e never terminates, that isn't relevant until you > actually evaluate e, which doesn't happen until you *call* (lambda (x) > (e x)). > > -- hendrik ____________________ Racket Users list: http://lists.racket-lang.org/users
