Re^2: The mystery of double negation

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Re^3: The mystery of double negation (bool 0) by tye (Sage) on Jul 12, 2015 at 22:51 UTC
The point of Double Negation is casting [...] any false value to 0. Nope, it's bitwise negation and doesn't work this way. It's basically - noop. Keeps value the same. Um: `% say "(!1)" % say "~~(!1)" 0` [download] So it could be an idiom that is applied to all Boolean values (so that Boolean false shows up as `'0'` instead of as the "invisible" `''`). It just does nothing when the Boolean value is already 1. - tye	[reply] [d/l] [select]
Re^4: The mystery of double negation (bool 0) by zwon (Abbot) on Jul 13, 2015 at 20:55 UTC
Interestingly, `!1` is `""`, `~~(!1)` is `0`, but `~~("")` is `""`. `$ perl -MO=Deparse -E'say !1; say ~~(!1); say ~~("");' use feature 'current_sub', 'evalbytes', 'fc', 'say', 'state', 'switch' +, 'unicode_strings', 'unicode_eval'; say !1; say 0; say ''; -e syntax OK` [download]	[reply] [d/l] [select]
Re^5: The mystery of double negation (SvPOK) by tye (Sage) on Jul 13, 2015 at 21:51 UTC
Of course. `!1` is a dual-value of empty string plus 0. Bitwise operators operate on integers if the scalar has an authoritative ("public") numeric value, otherwise they operate on strings. QED. - tye	[reply] [d/l]
Re^3: The mystery of double negation by shmem (Chancellor) on Jul 12, 2015 at 17:41 UTC
Right you are. Then... maybe chromatic can enlighten us? perl -le'print map{pack c,($-++?1:13)+ord}split//,ESEL'	[reply]
Re^3: The mystery of double negation by Laurent_R (Canon) on Jul 12, 2015 at 18:01 UTC
You're right, but, still, `~~1` will return 1, i.e a true value. Why the author of the module did that, I have no idea.	[reply] [d/l]


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