add1 comes for free, it's built-in. And it does exactly that!
[]'s Rodolfo Carvalho PS: please use this list [users at racket-lang.org] for user-related topics On Wed, Jun 1, 2011 at 02:14, Yingjian Ma <[email protected]> wrote: > Hi Rodolfo, > > Thank you for the help. ( I assume add1 is > > ( > define (add1 x) > > (+ x 1)) > > ) > > > On Tue, May 31, 2011 at 10:02 PM, Rodolfo Carvalho > <[email protected]>wrote: > >> [Moving the thread to the users mailing list] >> >> Yingjian, >> >> Good to see you persevering and getting it done! Congratulations. >> >> Before I reasoned about your code, just looking at its shape was enough to >> imagine it was not properly indented. >> You can use and abuse of DrRacket for your on good. >> >> Using the shortcut CTRL+I let's you reindent all of your code, which >> gives this: >> >> #lang racket >> (define (count-matches s l) >> (define (addone s l x) >> (cond >> [(empty? l) x] >> [(equal? s (first l)) (addone s (rest l) (+ 1 x))] >> [else (addone s (rest l) x)])) >> (addone s l 0)) >> >> >> The change in indentation makes clear that you're defining an auxiliary >> function and the result of calling count-matches is that of calling (addone >> s l 0). >> >> BTW "addone" doesn't really sound to me to describe what the function is >> doing -- it not always "adds one". >> >> >> Look how a similar implementation is possible without an inner auxiliary >> function: >> >> #lang racket >> (define (count-matches s l) >> (cond >> [(empty? l) 0] >> [(equal? s (first l)) (add1 (count-matches s (rest l)))] >> [else (count-matches s (rest l))])) >> >> (count-matches 'x '()) ; should be 0 >> (count-matches 'x '(a b x)) ; should be 1 >> (count-matches 'x '(x b x)) ; should be 2 >> >> >> []'s >> >> Rodolfo Carvalho >> >> >> >> On Wed, Jun 1, 2011 at 01:37, Yingjian Ma <[email protected]>wrote: >> >>> I finished it. The purpose of the function is to count the occurrance of >>> a letter in a list. Ex: >>> >>> (count-matches 'x '()) should be 0 >>> (count-matches 'x '(a b x)) should be 1 >>> (count-matches 'x '(x b x)) should be 2 >>> >>> The keywords I can use are limited. Thank you all for the help. >>> Another I need to write is to remove duplicated items from the list. >>> >>> It seems that under cond, one condition can only take one expression. >>> What can I do if I want to do two statements? >>> >>> Here is the code. >>> >>> (define (count-matches s l) >>> (define (addone s l x) >>> (cond >>> [(empty? l) x] >>> [(equal? s (first l)) (addone s (rest l) (+ 1 x))] >>> [else (addone s (rest l) x)])) >>> (addone s l 0)) >>> >>> >>> >> > > _________________________________________________ > For list-related administrative tasks: > http://lists.racket-lang.org/listinfo/dev >
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