The others have answered your explicit question, but one minor detail: Racket regards #f as the only false value. Everything else is true, which means that if there's a difference between these two tests:
(if foo "true" "false") (if (equal? #t foo) "true" "false") This can be relevant if you have functions with contracts that specify they will return a boolean. For example: (define (do-tests x) (unless (> x 7) (raise "x should be > 7")) (unless (< x 20) (raise "x should be less than 20")) (add1 x)) (displayln (if (do-tests 9) "Success!" "Failure!"))) This prints "Success!" Add a contract: (define/contract (do-tests x) (-> integer? boolean?) (unless (> x 7) (raise "x should be > 7")) (unless (< x 20) (raise "x should be less than 20")) x) (displayln (if (do-tests 9) "Success!" "Failure!"))) ...and it throws an exception saying "do-tests broke its own contract", because it returned an integer instead of either #t or #f. This is very rarely an issue, but if you absolutely positively must have a boolean value then you can do this: (not (not x)) which will force x to the boolean representation of its truth value -- that is, the value #f will continue to be #f and any other value will be converted to #t. On Sat, Apr 15, 2017 at 11:51 AM, 'William J. Bowman' via Racket Users <[email protected]> wrote: > On Sat, Apr 15, 2017 at 07:16:49AM -0700, Angus wrote: >> I have a function that can return either false or a node. I have to check >> four permutations: >> >> both nodes false, >> left node false, >> right node valid, left node false >> both node valid >> >> I am doing it using cond like this: >> >> (define (lines t) >> (cond [(and (false? (node-l t)) (false? (node-r t))) <<do no children >> thing>>] ;; if no children >> [(and (false? (node-l t)) (not (node-r t))) <<left child only>>] >> ;; if left child only >> [(and (node-l t) (false? (node-r t))) <<right child only>> ] ;; if >> right child only >> [else <<both children>> ])) ;; if both children > This looks like it might be a homework assignment. What Matthew said is true, > you can make this > shorter, but if your class encourages a specific explicit style then you > should be explicit instead. > >> the first cond line using and and false? works ok and the last cond line >> using else works ok. But how do I check for the middle two occurrences? > Any value that is not false in Racket is true. You can test for non-false > values using any conditional > construct, like `if` or `cond`. The operator `not` changes any non-false > value into false, and > false into true. > > > #lang racket > (define x 1) > (define y "Hello, World") > (define z false) > > (displayln x) > (displayln (not x)) > (displayln (not z)) > (displayln (not (not z))) > > (if x > (displayln x) > (displayln "This can't happen")) > > (if y > (displayln y) > (displayln "This also can't happen")) > > (cond > [(and x y (not z)) > (printf "~a and ~a and not ~a~n" x y z)] > [else (displayln "This cannot happen either")]) > > > -- > William J. Bowman > Northeastern University > College of Computer and Information Science > > -- > You received this message because you are subscribed to the Google Groups > "Racket Users" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Racket Users" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
