in reply to Converting to boolean
I use !!$x for this. Perl6 will give us ?$x, which ought to be clearer.
If I wanted to store the boolean back into the same variable, I might prefer a mutator: $x &&= 1.
However, there is a real difference:
sidhekin@blackbox:~$ perl my ($x, $y) = (0, 0); $x = $x && 1; # or $x &&= 1; $y = !!$y; print "\$x: '$x'\n\$y: '$y'\n"; __END__ $x: '0' $y: '' sidhekin@blackbox:~$
... something to be aware of :-)
print "Just another Perl ${\(trickster and hacker)},"
The Sidhekin proves Sidhe did it!
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