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? Additionally, chapter 8 is very interesting. I see that i can translate a simple recursive function the following way (define (sum n) (if (zero? n) 0 (+ n (sum (sub1 n))))) (sum 10) to (define (sum&co n k) (if zero? n) (k 0) (sum&co (sub1 n) (lambda (r) (k (+ n r)))))) (sum 10 (lambda (r) r)) but i really don't see why one would ever bother to do so. What are the benefits? the cost (hard to read / write / understand) seems high. thanks for your help (& further references appreciated). best regards mj
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