It shows 6 because a comma-separated literal list, when interpreted as a scalar, evaluates to the final value in the list, not the length of the list. If you evaluate "scalar(1,2,,4,,7)", it will equal 7. That's the comma operator at work-- even though it looks like a list (and both use commas), it's really just a sequential set of expressions.
Cheers, James On Wed, May 8, 2013 at 5:14 PM, <fe...@crowfix.com> wrote: > On Wed, May 08, 2013 at 02:04:37PM -0400, Derek Jones wrote: > > Huh? > > > > That produces 6 - as it should. What version of Perl are you working > with? > > > On May 8, 2013, at 1:58 PM, fe...@crowfix.com wrote: > > > > > scalar(1,2,,4,,6) > > You are right -- but this produces 4 first, as I had always thought it > would, and then 6, as I don't quite understand. I would never use > that kind of expression, and I sorta maybe understand why it says 6, > but only in hindsight. > > #!/usr/bin/perl > > my @ary = (1,2,,4,,6); > print "Step 1: ", scalar(@ary), "\n"; > print "Step 2: ", scalar(1,2,,4,,6), "\n"; > > -- > ... _._. ._ ._. . _._. ._. ___ .__ ._. . .__. ._ .. ._. > Felix Finch: scarecrow repairman & rocket surgeon / fe...@crowfix.com > GPG = E987 4493 C860 246C 3B1E 6477 7838 76E9 182E 8151 ITAR license > #4933 > I've found a solution to Fermat's Last Theorem but I see I've run out of > room o >
