[racket] [racket-dev] Please help me to fix the code

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.ma1955 at 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 <rhcarvalho at 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.ma1955 at 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
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.racket-lang.org/users/archive/attachments/20110601/7627e2d4/attachment.html>