`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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To po
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
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]
of course repeat works with functions:
user=> (take 4 (map #(% 4) (repeat inc)))
(5 5 5 5)
On Sun, Dec 23, 2012 at 11:56 PM, Marko Topolnik
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:
>>
>
*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]
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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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