On Sun, Jan 8, 2012 at 3:23 PM, Jordan Schatz <jor...@noionlabs.com> wrote: > This code runs, but I'm guessing that its not the "right way" to do it. > > (define (js-date [i (current-date)]) > (let ([original-format (date-display-format)] > [return ((λ () > (date-display-format 'rfc2822) > (date->string i #t)))]) > (date-display-format original-format) > return)) > > 1) In "some other language" using a function as the default value for an > argument is inefficient and frowned upon. Is that the case in racket?
Hi Jordan, Can you give an example of such a language? I'm curious. I'm not sure where the inefficiency would come from, unless computing the default value expression's value is costly. According to the documentation in: http://docs.racket-lang.org/reference/lambda.html#(form._((lib._racket/private/base..rkt)._lambda)) with regards to "default-expr": "... if no such argument is provided, the default-expr is evaluated to produce a value associated with id." From the reference docs, it sounds like that, unlike a language like Python, the default value is evaluated for every use of the function, rather than just once when the function's defined. We can experiment with this: ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; #lang racket (define (test [x (begin (printf "I'm evaluating at: ~a\n" (current-inexact-milliseconds)) 1)]) (add1 x)) ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; If you run test, you'll see that the default-expr is being evaluated for each call to test that depends on the default argument. If it is expensive to compute the default expression, we can bind a value to a variable, and then refer to that variable instead for the default expression. That way, the expression is just name lookup, which is constant-cost. Functions are named values, so I don't think they'd be particularly expensive to use as default expressions. ____________________ Racket Users list: http://lists.racket-lang.org/users
