He's repeating a function argument, not a function.
Meikel is correct that the second expression causes (some #{0} ...) to
return nil when number is not a multiple of 3 or 5, and then zero? fails.
As he suggests...
(reduce + (filter (fn [number] (some zero? (map mod (take 2 (repeat
number)) [3 5]))) (range 1 1000)))
...works (and returns 233168)
Sean
On Sun, Dec 23, 2012 at 11:56 PM, Marko Topolnik
<marko.topol...@gmail.com>wrote:
> *repeat* is not supposed to work with functions, but there's *repeatedly.*
>
>
> On Monday, December 24, 2012 4:20:23 AM UTC+1, Andrew Care wrote:
>>
>> I'm trying to use repeat with a function argument. This works:
>>
>> (reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
>> 9)) [3 5])))) (range 1 1000)))
>>
>> This doesn't:
>>
>> (reduce + (filter (fn [number] (zero? (some #{0} (map mod (take 2 (repeat
>> number)) [3 5])))) (range 1 1000)))
>>
>> Why can I use (repeat 9) and not (repeat number)?
>>
>
>
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