Re: $#="%c"; possible bug

[Abigail-II]

Setting $# sets the output format for printed numbers. I don't see why you expect '107' to be printed, I would expect 'k' to be printed.

But $# has been broken for eons, and it has been deprecated for even longer. I've never seen it used in code that wasn't intentionally obfuscated.

Abigail

[ambrus]

ord('k') is 107 so it should print 107, and the first statement should not affect it.

Is $# really broken for eons? Where's that from? See my reply for the other post Re: Re: $#="%c"; possible bug.

Update: effect->affect

[Abigail-II]

ord('k') is 107 so it should print 107

Uhm, no.

the first statement should not effect it.

Yes, it should. To quote myself: Setting $# sets the output format for printed numbers. 107 is still a number in my book.

From man perlvar:

       $#      The output format for printed numbers.  This vari­
               able is a half-hearted attempt to emulate awk's
               OFMT variable.  There are times, however, when awk
               and Perl have differing notions of what counts as
               numeric.  The initial value is "%.ng", where n is
               the value of the macro DBL_DIG from your system's
               float.h.  This is different from awk's default
               OFMT setting of "%.6g", so you need to set $#
               explicitly to get awk's value.  (Mnemonic: # is
               the number sign.)

               Use of $# is deprecated.

Is $# really broken for eons? Where's that from?

Yes it is. Your code is proof.

Abigail

[ambrus]

Good point. Now I see. I somehow thought $# only affects floating-point values. It seems that I didn't read the doc carefully enough.

And it seems that $# uses the system's builtin sprintf, not perl's.