I deleted my answer minutes after posting. The phrase "function argument"
is ambiguous ("argument to function", "argument which is a function") and I
jumped to the wrong conclusion.
On Monday, December 24, 2012 7:00:48 PM UTC+1, Sean Corfield wrote:
>
> 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.t...@gmail.com<javascript:>
> > 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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