Re: perl -e 'my $i = 0; $i = defined($i) ? (!!$i) : 0; print "i: $i\n";'

[hw]

X Dungeness wrote:

It's about what unary ! (bang operator) does to the operand

Here's the dissonance:

perl -E '$x=0; say "x=$x"; $x = !!$x; say "x=$x"'
x=0
x=

It behaves as you expect until you "bang" it twice.

I found a good explanation in the Camel:

"Unary ! performs logical negation, that is "not". The value
of a negated operand is true (1) if the operand is false
(numeric 0,string "0", the null string, or undefined) and
false ("") if the operand is true."

So double banging 0 yields "".

That doesn´t explain how and why a numerical value is converted into
a string.  It could be something else than the operator doing this.

In any case, which sane programmer would expect a bug like this.


perl -e 'my $i = 1; $i = defined($i) ? (!$i) : 0; use Data::Dumper; print 
Dumper($i);'
$VAR1 = '';


Logical negations do not involve turning numbers into strings.  If they
do, they are either something else or buggy.

How do you enforce a numerical context?

--
To unsubscribe, e-mail: beginners-unsubscr...@perl.org
For additional commands, e-mail: beginners-h...@perl.org
http://learn.perl.org/