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

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 <rhcarvalho at 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.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))
>>>>
>>>>
>>>>
>>>
>>
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