Re^2: return +0

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Re^3: return +0 by lidden (Curate) on Nov 27, 2007 at 09:07 UTC
In that example the `+` tells perl that the parens are not enclosing the arguments to a function call, so it is parsed correctly.	[reply] [d/l]
Re^4: return +0 by DrHyde (Prior) on Nov 27, 2007 at 10:36 UTC
Here's a nice simple example of when you might need to use it: `$ perl -e 'print (2+3)/5' 5` [download] How on earth does that print 5? Because it evaluates as "print the result of 2 + 3, then divide print's return value (which is 1 if print worked) by 5, and then do nothing with the result". This is obvious if you stick another print in there: `$ perl -e 'print print (2+3)/5' 50.2` [download] which is parsed as: `print((print(2+3))/5)`	[reply] [d/l] [select]
Re^5: return +0 by goibhniu (Hermit) on Nov 27, 2007 at 18:31 UTC
And just to close the logic loop (++DrHyde and lidden, btw), this leading plus helps the parser evaluate all those numbers into a single value before printing. `$ perl -e 'print +(2+3)/5' 1` [download] So I assume the author of the code the OP was reading thought there might be some ambiguity and wanted to help the parse out with that `return` statement. I humbly seek wisdom.	[reply] [d/l] [select]
Re^3: return +0 by syphilis (Archbishop) on Nov 27, 2007 at 08:50 UTC
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	[reply] [d/l]
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