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

Hello Monks
Can anybody explan me why the following code prints '1'?
my $a = 1;
my $b = 0;
my $c = $a and $b;
print $c;

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did i have to use different literals for boolean algebra? but even then writing 
my $a = (1==1);
my $b = (1==0);
my $c = $a and $b;
print $c;

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the result is 1. but (1==0) evaluates to an empty string

please give me some clue
Thanks

Replies are listed 'Best First'.
Re: strange boolean algebra behaviour
by moritz (Cardinal) on Apr 04, 2012 at 10:14 UTC

    The problem is that and has very loose precedence, looser than the assignment operator. So you need to write $a && $b or ($a and $b). See perlop for more details.

    In Perl 5, the false value is represented as an empty string that can numify to 0 without a warning. That's what you are seeing in your second attempt.

    Perl 6 - second systems done right
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      aarrrgghh!!!
      I could smash my head on table.
      operator precedence is a thing that i should have remember.

      I was utterly convinced that '&&' and 'and' are synonyms. Thank you.
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        I could smash my head on table.

        Don't do that, that could hurt.

        But if you do do it, make/post a video

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      use strict;
use warnings;

my $a = 1;
my $b = 0;
my $c = $a and $b;
print "c=$c\n";

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      This gives me warning that "Useless use of private variable in void context at tmp16.pl line 6.". line 6 is "my $c = $a and $b;". I first tried -MO=Deparse. 
      C:\temp>perl -MO=Deparse tmp16.pl
Useless use of private variable in void context at tmp16.pl line 6.
use warnings;
use strict 'refs';
my $a = 1;
my $b = 0;
$b if my $c = $a;
print "c=$c\n";
$a = 1;
$b = 0;
$c = $a && $b;
print "c=$c\n";
tmp16.pl syntax OK

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      So, Deparse maybe says $b of "$b if my $c = $a;" is the useless use of private variable.
      regards.

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        -p 

        $ perl -MO=Deparse -e " my $foo = $bar and $baz; "
$baz if my $foo = $bar;
-e syntax OK

$ perl -MO=Deparse,-p -e " my $foo = $bar and $baz; "
((my $foo = $bar) and $baz);
-e syntax OK

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