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!