I suppose I should ask this in a Ruby forum, but since I'm so used to slinging other languages here ...
To find the Nth root of number is simple: raise the number to the reciprocal of N. For example, to find the cube root of 8:
$ perl -le 'print 8 ** (1/3)' 2
But you can't quite do that in Ruby:
$ ruby -e 'puts 8 ** (1/3)' 1
But this is a "feature", not a bug (*cough*) because the 1/3 is considered integer math and evaluates to 0, leaving you with 8 to the 0th power. Anything raised to the power of 0 results in 1. So far so good.
So to force floating point math, use a floating point number:
$ ruby -e 'puts 8 ** (1/3.0)' 2
And all is good. Except ...
Let's take the square root of 1:
$ ruby -e 'puts 1 ** (1/2.0)' 1.0
Now let's take the square root of -1:
$ ruby -e 'puts -1 ** (1/2.0)' -1.0
Huh? The square root of -1 is imaginary (or i, if you want to be specific). What's going on here?
Yes, I know about Math.sqrt, which at least thoughtfully throws an exception rather than give an incorrect value:
$ ruby -e 'puts Math.sqrt(-1)' -e:1:in `sqrt': Numerical argument out of domain - sqrt (Errno::EDOM) from -e:1
Re:Perl is better?
Burak on 2009-02-21T01:07:07
indeed you are:
C:\Users\burak>perl -MO=Deparse -le "print (-1) ** (1/2)"
BEGIN { $/ = "\n"; $\ = "\n"; }
print(-1) ** 0.5;
-e syntax OKtry this instead:
print( (-1) ** (1/2) )
Re:Perl is better?
frew on 2009-02-21T01:13:36
Hah! The funny thing is that I was talking to someone about this bug YESTERDAY.
Thanks.Re:Perl is better?
garu on 2009-02-21T03:10:48
Ruby is no worse either, then:
$ ruby -e 'puts ((-1) ** (1/2.0))'
NaNRe:Perl is better?
drhyde on 2009-02-23T12:21:17
Having to liberally splatter brackets around the place looks no better than having to force something to not be interpreted as an integer. It's certainly just as obscure.
