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• Practical Scheme Programming

  - by Ixmatus

  It's been a few months since I've touched Scheme and decided to implement a command line income partitioner using Scheme. My initial implementation used plain recursion over the continuation, but I figured a continuation would be more appropriate to this type of program. I would appreciate it if anyone (more skilled with Scheme than I) could take a look at this and suggest improvements. I'm that the multiple (display... lines is an ideal opportunity to use a macro as well (I just haven't gotten to macros yet). (define (ab-income) (call/cc (lambda (cc) (let ((out (display "Income: ")) (income (string->number (read-line)))) (cond ((<= income 600) (display (format "Please enter an amount greater than $600.00~n~n")) (cc (ab-income))) (else (let ((bills (* (/ 30 100) income)) (taxes (* (/ 20 100) income)) (savings (* (/ 10 100) income)) (checking (* (/ 40 100) income))) (display (format "~nDeduct for bills:---------------------- $~a~n" (real->decimal-string bills 2))) (display (format "Deduct for taxes:---------------------- $~a~n" (real->decimal-string taxes 2))) (display (format "Deduct for savings:-------------------- $~a~n" (real->decimal-string savings 2))) (display (format "Remainder for checking:---------------- $~a~n" (real->decimal-string checking 2)))))))))) Invoking (ab-income) asks for input and if anything below 600 is provided it (from my understanding) returns (ab-income) at the current-continuation. My first implementation (as I said earlier) used plain-jane recursion. It wasn't bad at all either but I figured every return call to (ab-income) if the value was below 600 kept expanding the function. (please correct me if that apprehension is incorrect!)

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• Scheme: what are the benefits of letrec?

  - by Ixmatus

  While reading "The Seasoned Schemer" I've begun to learn about letrec. I understand what it does (can be duplicated with a Y-Combinator) but the book is using it in lieu of recurring on the already defined function operating on arguments that remain static. An example of an old function using the defined function recurring on itself (nothing special): (define (substitute new old lat) (cond ((null? l) '()) ((eq? (car l) old) (cons new (substitute new old (cdr l)))) (else (cons (car l) (substitute new old (cdr l)))))) Now for an example of that same function but using letrec: (define (substitute new old lat) (letrec ((replace (lambda (l) (cond ((null? l) '()) ((eq? (car l) old) (cons new (replace (cdr l)))) (else (cons (car l) (replace (cdr l)))))))) (replace lat))) Aside from being slightly longer and more difficult to read I don't know why they are rewriting functions in the book to use letrec. Is there a speed enhancement when recurring over a static variable this way because you don't keep passing it?? Is this standard practice for functions with arguments that remain static but one argument that is reduced (such as recurring down the elements of a list)? Some input from more experienced Schemers/LISPers would help!

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