I see. I cannot use it as I said in my first post. But I got your idea. I'll rename addone to count or something else.
On Tue, May 31, 2011 at 10:29 PM, Rodolfo Carvalho <rhcarva...@gmail.com>wrote: > 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 <yingjian.ma1...@gmail.com>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 >> <rhcarva...@gmail.com>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 <yingjian.ma1...@gmail.com>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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