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

 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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