Re^3: return +0

I belive '+' has some importance.

For me, return values of +0, -0, and 0 are indistinguishable from each other - which means that the '+' has no significance at all:

C:\_32>perl -MDevel::Peek -e "Dump(+0)"
SV = IV(0x1fe2d28) at 0x343cbc
  REFCNT = 1
  FLAGS = (PADBUSY,PADTMP,IOK,READONLY,pIOK)
  IV = 0

C:\_32>perl -MDevel::Peek -e "Dump(-0)"
SV = IV(0x432d2c) at 0xd73cec
  REFCNT = 1
  FLAGS = (PADBUSY,PADTMP,IOK,READONLY,pIOK)
  IV = 0

C:\_32>perl -MDevel::Peek -e "Dump(0)"
SV = IV(0x2492d28) at 0x23cbc
  REFCNT = 1
  FLAGS = (PADBUSY,PADTMP,IOK,READONLY,pIOK)
  IV = 0
[download]

(Replace the double quotes with single quotes if you're on a *nix-type operating system.)

Cheers,
Rob

Comment on Re^3: return +0 Download Code

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Re^4: return +0 by diotalevi (Canon) on Nov 27, 2007 at 16:53 UTC
Just note that -.0 is distinct from .0. ⠤⠤ ⠙⠊⠕⠞⠁⠇⠑⠧⠊	[reply]
Re^5: return +0 (-.0?) by tye (Sage) on Nov 27, 2007 at 22:10 UTC
Perhaps you could be less vague. There is no such distinction in my tests: `> perl -MDevel::Peek -e"Dump(+.0)" SV = NV(0xa4a5fc) at 0x3d53c4 REFCNT = 1 FLAGS = (PADBUSY,PADTMP,NOK,READONLY,pNOK) NV = 0 > perl -MDevel::Peek -e"Dump(-.0)" SV = NV(0xa4a60c) at 0x3d53f4 REFCNT = 1 FLAGS = (PADBUSY,PADTMP,NOK,READONLY,pNOK) NV = 0 > perl -MDevel::Peek -e"Dump(.0)" SV = NV(0xa4a5fc) at 0x3d53c4 REFCNT = 1 FLAGS = (PADBUSY,PADTMP,NOK,READONLY,pNOK) NV = 0` [download] - tye	[reply] [d/l]
Re^6: return +0 (-.0?) by Sidhekin (Priest) on Nov 28, 2007 at 22:23 UTC
It depends on the math libraries? They're different here, at least: `sidhekin@blackbox[23:22:41]~$ perl -MDevel::Peek -le 'Dump(.0)' SV = NV(0x816e490) at 0x814dce4 REFCNT = 1 FLAGS = (PADBUSY,PADTMP,NOK,READONLY,pNOK) NV = 0 sidhekin@blackbox[23:22:51]~$ perl -MDevel::Peek -le 'Dump(-.0)' SV = NV(0x816e4a0) at 0x814dd20 REFCNT = 1 FLAGS = (PADBUSY,PADTMP,NOK,READONLY,pNOK) NV = -0 sidhekin@blackbox[23:22:53]~$` [download] `print "Just another Perl ${\(trickster and hacker)},"` The Sidhekin proves Sidhe did it!	[reply] [d/l] [select]