From: Chas. Owens
Sent: Friday, July 07, 2017 12:34 AM
To: hw ; beginners@perl.org
Subject: Re: perl -e 'my $i = 0; $i = defined($i) ? (!!$i) : 0; print "i:
$i\n";'
On Thu, Jul 6, 2017 at 9:38 AM hw <h...@gc-24.de> wrote:
Chas. Owens wrote:
$i started off as an IV, but gets promoted to a PVIV by being used in
string context.
I find it a little surprising that use of the '!' operator is all that's
needed to add the stringification stuff:
C:\>perl -MDevel::Peek -le "$i=0; Dump($i); Dump(!$i);"
SV = IV(0x93ece0) at 0x93ece4
REFCNT = 1
FLAGS = (IOK,pIOK)
IV = 0
SV = PVNV(0xb0a934) at 0xb03090
REFCNT = 2147483644
FLAGS = (IOK,NOK,POK,READONLY,pIOK,pNOK,pPOK)
IV = 1
NV = 1
PV = 0xb0330c "1"\0
CUR = 1
LEN = 12
If the '!' operator didn't do that, then I believe the OP would be seeing
precisely what he expects.
So ... why should the '!' operator *not* respect the string/numeric context
of the operand ?
Why does it insist on logically negating *both* the numeric and string slots
of the variable, even when those slots are initially unset ?
Maybe it's just that that's the way it has always been done - or maybe
there's a more convincing reason ?
Cheers,
Rob
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