CODEWITHSUNDEEP
perloperatorswarnings
sid_com
sid_com

Reputation: 25117

Warnings on equality operators

Has something changed in Perl or has it always been this way, that examples like the second ($number eq 'a') don't throw a warning?

#!/usr/bin/env perl
use warnings;
use 5.12.0;

my $string = 'l';
if ($string == 0) {};

my $number = 1;
if ($number eq 'a') {};


# Argument "l" isn't numeric in numeric eq (==) at ./perl.pl line 6.

Upvotes: 3

Views: 114

Answers (3)

ikegami
ikegami

Reputation: 385657

Did you use a lowercase "L" on purpose? It's often hard to tell the difference between a lowercase "L" and one. You would have answered your own question if you had used a one instead.

>perl -wE"say '1' == 0;"


>perl -wE"say 1 eq 'a';"


>

As you can see,

  • If one needs a number, Perl will convert a string to a number without warning.
  • If one needs a string, Perl will convert a number to a string without warning.

Very consistent.

You get a warning when you try to convert a lowercase L to a number, but how is that surprising?

Upvotes: 0

Borodin
Borodin

Reputation: 126722

Perl will be try to convert a scalar to the type required by the context where it is used.

There is a valid conversion from any scalar type to a string, so this is always done silently.

Conversion to a number is also done silently if the string passes a looks_like_number test (accessible through Scalar::Util). Otherwise a warning is raised and a 'best guess' approximation is done anyway.

my $string = '9';
if ( $string == 9 ) { print "YES" };

Converts the string silently to integer 9, the test succeeds and YES is printed.

my $string = '9,8';
if ( $string == 9 ) { print "YES" };

Raises the warning Argument "9,8" isn't numeric in numeric eq (==), converts the string to integer 9, the test succeeds and YES is printed.

To my knowledge it has always been this way, at least since v5.0.

Upvotes: 6

Alan Haggai Alavi
Alan Haggai Alavi

Reputation: 74222

It has been that way.

In the first if, l is considered to be in numeric context. However, l cannot be converted to a number. Therefore, a warning is emitted.

In the second if, the number 1 is considered to be in string context. Therefore the number 1 is converted to the string '1' before comparison and hence no warnings are emitted.

Upvotes: 4

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