How do I tell if a variable has a numeric value in Perl?

85

Is there a simple way in Perl that will allow me to determine if a given variable is numeric? Something along the lines of:

if (is_number($x))
{ ... }

would be ideal. A technique that won't throw warnings when the -w switch is being used is certainly preferred.

This question is tagged with perl numbers

~ Asked on 2008-08-15 19:43:15

14 Answers


130

Use Scalar::Util::looks_like_number() which uses the internal Perl C API's looks_like_number() function, which is probably the most efficient way to do this. Note that the strings "inf" and "infinity" are treated as numbers.

Example:

#!/usr/bin/perl

use warnings;
use strict;

use Scalar::Util qw(looks_like_number);

my @exprs = qw(1 5.25 0.001 1.3e8 foo bar 1dd inf infinity);

foreach my $expr (@exprs) {
    print "$expr is", looks_like_number($expr) ? '' : ' not', " a number\n";
}

Gives this output:

1 is a number
5.25 is a number
0.001 is a number
1.3e8 is a number
foo is not a number
bar is not a number
1dd is not a number
inf is a number
infinity is a number

See also:

~ Answered on 2008-08-26 16:53:34


23

The original question was how to tell if a variable was numeric, not if it "has a numeric value".

There are a few operators that have separate modes of operation for numeric and string operands, where "numeric" means anything that was originally a number or was ever used in a numeric context (e.g. in $x = "123"; 0+$x, before the addition, $x is a string, afterwards it is considered numeric).

One way to tell is this:

if ( length( do { no warnings "numeric"; $x & "" } ) ) {
    print "$x is numeric\n";
}

If the bitwise feature is enabled, that makes & only a numeric operator and adds a separate string &. operator, you must disable it:

if ( length( do { no if $] >= 5.022, "feature", "bitwise"; no warnings "numeric"; $x & "" } ) ) {
    print "$x is numeric\n";
}

(bitwise is available in perl 5.022 and above, and enabled by default if you use 5.028; or above.)

~ Answered on 2010-09-27 17:16:12


23

Check out the CPAN module Regexp::Common. I think it does exactly what you need and handles all the edge cases (e.g. real numbers, scientific notation, etc). e.g.

use Regexp::Common;
if ($var =~ /$RE{num}{real}/) { print q{a number}; }

~ Answered on 2008-08-15 21:01:16


9

Usually number validation is done with regular expressions. This code will determine if something is numeric as well as check for undefined variables as to not throw warnings:

sub is_integer {
   defined $_[0] && $_[0] =~ /^[+-]?\d+$/;
}

sub is_float {
   defined $_[0] && $_[0] =~ /^[+-]?\d+(\.\d+)?$/;
}

Here's some reading material you should look at.

~ Answered on 2008-08-15 19:58:52


9

A simple (and maybe simplistic) answer to the question is the content of $x numeric is the following:

if ($x  eq  $x+0) { .... }

It does a textual comparison of the original $x with the $x converted to a numeric value.

~ Answered on 2012-10-17 15:10:45


4

Not perfect, but you can use a regex:

sub isnumber 
{
    shift =~ /^-?\d+\.?\d*$/;
}

~ Answered on 2008-08-15 19:49:44


3

I don't believe there is anything builtin to do it. For more than you ever wanted to see on the subject, see Perlmonks on Detecting Numeric

~ Answered on 2008-08-15 20:44:02


2

rexep not perfect... this is:

use Try::Tiny;

sub is_numeric {
  my ($x) = @_;
  my $numeric = 1;
  try {
    use warnings FATAL => qw/numeric/;
    0 + $x;
  }
  catch {
    $numeric = 0;
  };
  return $numeric;
}

~ Answered on 2010-12-10 18:58:12


2

A slightly more robust regex can be found in Regexp::Common.

It sounds like you want to know if Perl thinks a variable is numeric. Here's a function that traps that warning:

sub is_number{
  my $n = shift;
  my $ret = 1;
  $SIG{"__WARN__"} = sub {$ret = 0};
  eval { my $x = $n + 1 };
  return $ret
}

Another option is to turn off the warning locally:

{
  no warnings "numeric"; # Ignore "isn't numeric" warning
  ...                    # Use a variable that might not be numeric
}

Note that non-numeric variables will be silently converted to 0, which is probably what you wanted anyway.

~ Answered on 2008-08-15 20:56:36


1

I found this interesting though

if ( $value + 0 eq $value) {
    # A number
    push @args, $value;
} else {
    # A string
    push @args, "'$value'";
}

~ Answered on 2015-10-01 14:15:28


1

Try this:

If (($x !~ /\D/) && ($x ne "")) { ... }

~ Answered on 2013-05-31 16:01:54


0

You can use Regular Expressions to determine if $foo is a number (or not).

Take a look here: How do I determine whether a scalar is a number

~ Answered on 2016-02-26 16:55:43


0

Personally I think that the way to go is to rely on Perl's internal context to make the solution bullet-proof. A good regexp could match all the valid numeric values and none of the non-numeric ones (or vice versa), but as there is a way of employing the same logic the interpreter is using it should be safer to rely on that directly.

As I tend to run my scripts with -w, I had to combine the idea of comparing the result of "value plus zero" to the original value with the no warnings based approach of @ysth:

do { 
    no warnings "numeric";
    if ($x + 0 ne $x) { return "not numeric"; } else { return "numeric"; }
}

~ Answered on 2015-12-31 13:47:24


-1

if ( defined $x && $x !~ m/\D/ ) {} or $x = 0 if ! $x; if ( $x !~ m/\D/) {}

This is a slight variation on Veekay's answer but let me explain my reasoning for the change.

Performing a regex on an undefined value will cause error spew and will cause the code to exit in many if not most environments. Testing if the value is defined or setting a default case like i did in the alternative example before running the expression will, at a minimum, save your error log.

~ Answered on 2013-09-21 16:14:58


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