`repeat` takes a value and returns a sequence of that value.
There's another function, `repeatedly`, which takes a function and returns
a sequence of values returned by calling that function.
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>>
>>
>> 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
>>> (re
>> 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
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:
>>
>> (reduc
*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
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Hi,
Am 24.12.12 04:20, schrieb Andrew Care:
> Why can I use (repeat 9) and not (repeat number)?
Because you call zero? on nil, which doesn't work. Instead of (zero?
(some #{0} ...)) use (some zero? ...).
Kind regards
Meikel
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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] (ra
