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How do I decide if a variable is numeric in Perl?

Answer 1

+6 A:

if (/\D/)            { print "has nondigits\n" }
if (/^\d+$/)         { print "is a whole number\n" }
if (/^-?\d+$/)       { print "is an integer\n" }
if (/^[+-]?\d+$/)    { print "is a +/- integer\n" }
if (/^-?\d+\.?\d*$/) { print "is a real number\n" }
if (/^-?(?:\d+(?:\.\d*)?&\.\d+)$/) { print "is a decimal number\n" }
if (/^([+-]?)(?=\d&\.\d)\d*(\.\d*)?([Ee]([+-]?\d+))?$/)
                     { print "a C float\n" }

taken from here: http://rosettacode.org/wiki/Determine_if_a_string_is_numeric#Perl

Parkyprg 2010-09-27 12:45:33

If you insist on a regex, get it from Regexp::Common :)

brian d foy 2010-09-27 15:33:24

And, every one of these regexes that uses the $ anchor allows a newline at the end of the value. That's a mistake in perlfaq4 (I've now patched the docs to remove that mistake), where Rosetta Code got its example. This example, however, was to determine what sort of number it was, not that it was any sort of number.

brian d foy 2010-09-27 15:41:52

Answer 2

+1 A:

using regex, its good to use:

if ($var =~ /^\d+?$/) {
    // integer
}

Alternatively, you can use POSIX.

use POSIX;

if (isdigit($var)) {
    // integer
}

Ruel 2010-09-27 12:49:25

This just finds digits. It misses literals that are numbers that have more than mere digits.

brian d foy 2010-09-27 15:34:07

Answer 3

+13 A:

You can use the looks_like_number() function from the core Scalar::Util module.
See also the question in perlfaq: How do I determine whether a scalar is a number/whole/integer/float?

eugene y 2010-09-27 12:50:21

This is arguably the most correct answer, since `Scalar::Util` hooks into the Perl API to actually ask Perl whether it thinks the variable looks numeric. Thus, for the common task of avoiding 'numeric' warnings, this will always work. Of course, sometimes it's better to simply say `no warnings 'numeric';`...

Adam Bellaire 2010-09-27 20:32:14

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How do I decide if a variable is numeric in Perl?

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