Re: Is the number before, in or after the interval? using spaceship operator <=>

> It can be find using two basic comparisons.

Strange, I count 6 operations! :)

 floor (($i - $l ) / ($r - $l +1 )) <=> 0
   1        2      3     4    5      6
[download]

Cheers Rolf

_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}

PS: Please don't post such stuff in SOPW without a clear question, you are keeping others busy moderating that stuff:

Tricks belong to Cool Uses For Perl

Comment on Re: Is the number before, in or after the interval? using spaceship operator <=> Download Code

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Re^2: Is the number before, in or after the interval? using spaceship operator <=> by rsFalse (Chaplain) on Nov 21, 2014 at 14:43 UTC
Hm.. yes. And that is simpler solution with two basic comparisons: `map { ("-1", " 0", " 1")[ ($_ >= $l) + ($_ > $r) ] } -9 .. 10;`	[reply] [d/l]
Re^3: Is the number before, in or after the interval? using spaceship operator <=> by LanX (Saint) on Nov 21, 2014 at 15:23 UTC
`($_ > $r) - ($_ < $l)` Should already do (can't test on Android :) update tested! =) `DB<41> sub tst { ($_ > $r) - ($_ < $l) } DB<42> $l=3;$r=6 => 6 DB<43> print tst(),"\t" for 0..9 => "" -1 -1 -1 0 0 0 0 1 1 1` [download] Cheers Rolf _{(addicted to the Perl Programming Language and ☆☆☆☆ :)}	[reply] [d/l] [select]