Float regex matching

Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

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Re: Float regex matching by davido (Cardinal) on Feb 24, 2016 at 16:21 UTC
Take this example: `perl -E 'say "match" if 0.0 =~ m/^\d+/'` [download] This matches. But "`.`" is not included in `\d`, so why does it match? Because 0.0 is a number that gets implicitly stringified for the purpose of binding to a pattern match. The rules of numeric stringification are to evaluate the numeric value and convert that numeric value to a string representation. Try this, to see it at work: `perl -E 'my @match = 0.0 =~ m/^(\d+)$/; say "@match";'` [download] The output will be "`0`", demonstrating that the value is what gets stringified, not the source code you typed into the keyboard to create the value. When you wrap the `0.0` in quotes first, you create a string, "`0.0`", which is taken exactly as it is, for the purpose being bound to a pattern match. On the other hand, the value 0.5 stringifies as "`0.5`", preserving the decimal point in the process. The decimal doesn't match `\d`, so the match fails. Dave	[reply] [d/l] [select]
Re: Float regex matching by Your Mother (Archbishop) on Feb 24, 2016 at 16:09 UTC
I expect you can compare these two to see. `perl -le 'print "OK" if 0.0 =~ /^\d+$/' perl -le 'print "OK" if "0.0" =~ /^\d+$/'` [download]	[reply] [d/l]
Re^2: Float regex matching by Anonymous Monk on Feb 24, 2016 at 16:21 UTC
Looks like Perl resolves 0.0 to 0	[reply]
Re^3: Float regex matching by AnomalousMonk (Archbishop) on Feb 24, 2016 at 16:58 UTC
... Perl resolves 0.0 to 0 As davido points out below, Perl stringizes 0.0 to '0': `c:\@Work\Perl\monks>perl -wMstrict -le "my $f = 0.0; my $fs = qq{$f}; print qq{stringization of 0.0: '$fs'}; ;; $f = 0.5; $fs = qq{$f}; print qq{stringization of 0.5: '$fs'}; " stringization of 0.0: '0' stringization of 0.5: '0.5'` [download] Give a man a fish: `<%-{-{-{-<`	[reply] [d/l] [select]