Shlomi Fish wrote:
Hi hw!
Please see
http://www.shlomifish.org/philosophy/computers/netiquette/email/reply-to-list.html
.
On Sat, 1 Jul 2017 19:15:22 +0200 hw <h...@gc-24.de> wrote:
Shlomi Fish wrote:
Hi Shawn!
On Sat, 1 Jul 2017 11:32:30 -0400
Shawn H Corey <shawnhco...@gmail.com> wrote:
On Sat, 1 Jul 2017 17:27:02 +0200
hw <h...@gc-24.de> wrote:
Hi,
can someone please explain this:
perl -e 'my $i = 0; $i = defined($i) ? (!!$i) : 0; print "i: $i\n";'
i:
Particularly:
+ Why doesn´t it print 1?
Because !!$i is zero
+ How is this not a bug?
Nope, no bug.
+ What is being printed here?
!!$i which is !(!(0)) which is !(1) which is 0
I suspect !1 returns an empty string in scalar context.
What would be the reasoning for a numerical operator turning
numerical values into (empty) strings within a numerical context?
"!" is not a numerical operator - it is a *logical* operator. If one passes a
true value to it one gets a false value.
False and true are genuinely numeric. You can´t say for a string
whether it is true or false; it is a string.
Else, one gets a true value. See
http://perldoc.perl.org/perlop.html . If used in a numerical context, false
values are treated as zeroes, but not all of them are zeroes in other contexts.
There is no other context involved here than a numerical one. There
is also no reason to implicitly change the context to any other one,
like a string context.
perl -e 'my $i = 0; $i = defined($i) ? (!!$i) : 0; print $i;'
perl -e 'my $i = 0; $i = defined($i) ? (!!$i) : 0; printf("%d", $i);'
perl -e 'my $i = undef; $i = defined($i) ? (!!$i) : 0; print $i;'
These give you different results, and that is just wrong. I did
assign a /number/ to $i and never a string.
Regards,
Shlomi
+ How do you do what I intended in perl?
How do we know what you intend?
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