Question

I have a condition:

next if ( ! ($x or $y or $z) );

The logic of the check is that at least one must be numerically non-zero to continue in the loop.

I trust that they actually are numbers.

The problem is that perl stores floats as strings internally. So a check on ! $x where $x='0.00' does not actually evaluate to true: my $x = 0.00; if ( ! $x ) { never_gets_here(); }

What is the easiest way to force numeric evaluation of a variable without making the line too verbose?

Answer 1

I’m not sure where you get the idea that Perl stores floats as strings. Floats and strings are different things:

perl -le 'print 1 if 0.00'
perl -le 'print 2 if "0.00"'
2

If you want to force numeric context on an unknown scalar, you can just add zero to it, e.g.

unless ( $x + 0 ) { ... }