Do you know that...
Yes, it's documented in perlapi. I didn't assume that would be a concern for the OP.
The old purePerl version of Scalar::Util has a purePerl version of looks_like_number, which includes the following code:
return 1
if ( $] >= 5.008 and /^(Inf(inity)?NaN)$/i )
or ( $] >= 5.006001 and /^Inf$/i );
So if we wanted to eliminate that behavior, we could either copy/paste and modify the entire purePerl version from an old version of Scalar::Util (but it has issues with '0 but true'), or use the modern version of Scalar::Util::looks_like_number like this:
sub looks_like_finite_number {
my $candidate = shift;
return 0 if ! looks_like_number( $candidate );
return 0
if ( $] >= 5.008 and $candidate =~ /^(Inf(inity)?NaN)$/i)
or ( $] >= 5.006001 and $candidate =~ /^Inf$/i );
return 1;
}
Or, in perlfaq4 there is a series of regexes given for determining if one is looking at a number under the section, "How do I determine whether a scalar is a number/whole/integer/float? It's not pretty, and that FAQ answer has even changed over the years; it used to use a bunch of "if" statements. Now it uses given/when. *sigh*
