It is the exponentiation operator:
$ perl -E 'say 3 ** 2;'
9
Cheers,
Laurent.
Le mer. 3 nov. 2021 Γ 20:05, Wes Peng <w...@coakmail.com> a Γ©crit :
> What's the "**" operator in perl? I most of the time use R for math, not
> familiar with this operation.
>
> Thanks
>
> On Sun, Oct 31, 2021 at 3:38 AM Sean McAfee <eef...@gmail.com> wrote:
>
>> Recently I was golfing the "hyperfactorial," defined for a number π as
>> π**π Γ (π-1)**(π-1) Γ (π-2)**(π-2) Γ ... Γ 1. I created a quite
>> concise Raku function:
>>
>> { [*] [\*] $_...1 }
>>
>> The only problem was that this function returns zero for a zero input,
>> whereas the hyperfactorial of 0 is supposed to be 1. Of course I could
>> have just handled zero as a special case, but I hoped to find something
>> comparably short. After a bit of thought I tried reversing both the range
>> and the operator:
>>
>> { [*] [\R*] 1..$_ }
>>
>> It worked! But I couldn't quite see how. * is commutative, so isn't it
>> exactly the same as R*?
>>
>> It turns out, in Raku it isn't quite the same. On the operators page, I
>> found that the R metaoperator produces an operator that reverses the order
>> of the arguments, but *also* has the opposite associativity. So, for
>> example, [\R*] 1..4 reduces from the right, producing the list (4, 12, 24,
>> 24). Somehow I had formed the idea that Raku operators are left-, right-,
>> or list-associative, but I was wrong.
>>
>> Anyway, pretty cool!
>>
>>
