Re: How to check if a scalar value is numeric or string?

I tried to do it using regular expressions but I couldn't have a clear shot, there is another feature, use the complement operator, In order to determine if a variable is a number or a string, employ the principle that &ing a variable with its complement should give 0 as a result if that variable holds a number, whereas it is not the case when that variable is a string.

try extending on this principle, here is a snippet:

my $x=111;
my $y="111";

print "\$x is a number\n" if !($x & ~$x);
print "\$y is a string\n" if ($y & ~$y);
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Re^2: How to check if a scalar value is numeric or string? by ikegami (Patriarch) on Aug 27, 2009 at 16:18 UTC
By the way, `$x & ~$x` can also be written as `$x ^ $x`.	[reply] [d/l] [select]
Re^3: How to check if a scalar value is numeric or string? by Anonymous Monk on Feb 07, 2017 at 19:42 UTC
As of perl 5.24 you can no longer use this trick because ^ does not like strings with code points over 0xFF. Right now it is deprecated, but will presumably be fatal later.	[reply]
Re^4: How to check if a scalar value is numeric or string? by voidzero (Initiate) on Jul 18, 2023 at 08:01 UTC
Hi, this is day 2 of my perl journey and I'm doing some exercises in Learning Perl. Where can I read more on what `^` was, why it was deprecated, and on `$x & ~$x`?	[reply] [d/l] [select]
Re^5: How to check if a scalar value is numeric or string? by hippo (Archbishop) on Jul 18, 2023 at 08:59 UTC
Re^5: How to check if a scalar value is numeric or string? by eyepopslikeamosquito (Archbishop) on Jul 18, 2023 at 09:45 UTC
Re^5: How to check if a scalar value is numeric or string? by haj (Vicar) on Jul 18, 2023 at 08:57 UTC
Re^2: How to check if a scalar value is numeric or string? by bgupta (Novice) on Aug 27, 2009 at 16:07 UTC
Many thanks, you saved my day, this approach works perfectly for me.	[reply]
Re^3: How to check if a scalar value is numeric or string? by ig (Vicar) on Aug 28, 2009 at 00:55 UTC
But watch out for side effects. `my $x = 1; my $y = '1'; print "x is a ", (($x ^ $x)?"string":"number"), "\n"; print "y is a ", (($y ^ $y)?"string":"number"), "\n"; print "\$x = $x\n"; print "\$y = $y\n"; print "x is a ", (($x ^ $x)?"string":"number"), "\n"; print "y is a ", (($y ^ $y)?"string":"number"), "\n"; my $z = $x + $y; print "\$z = $z\n"; print "x is a ", (($x ^ $x)?"string":"number"), "\n"; print "y is a ", (($y ^ $y)?"string":"number"), "\n";` [download] gives `x is a number y is a string $x = 1 $y = 1 x is a number y is a string $z = 2 x is a number y is a number` [download] It might be better to revise your program so that it doesn't depend on such a subtle distinction.	[reply] [d/l] [select]