Undef value assigned to variable. - perl

Question

The return value of 5 > 6 is undef. However when I assign the value of 5 > 6 to a variable that variable is defined. How do I pass the value of a statement using a comparison operator that resolves to undef on failure?
#!/usr/bin/perl
use strict ;
use warnings;
print'Five is more than six ? ', 5 > 6, "\n";
print 'Five is less than six ? ' , 5 < 6 , "\n";
my $wiz = 5 > 6 ;
if (defined($wiz)) {
print '$wiz is defined' ;
} else {
print '$wiz is undefined' ;
}
$ ./lessthan
Five is more than six ?
Five is less than six ? 1
$wiz is defined

Answer 1

How do I pass the value of a statement using a comparison operator that resolves to undef on failure?
Generally speaking, Perl doesn't promise to return any specific true or false value from its operators, and < is no exception. If you want specific values, you will need something like
$boolean ? $value_for_true : $value_for_false
so
my $wiz = 5 > 6 ? 1 : undef;
If you only care about the value you get for false, you could also use the following:
my $wiz = 5 > 6 || undef;
The two options turn out to be equivalent, but this isn't guaranteed.
The return value of 5 > 6 is undef. However when I assign the value of 5 > 6 to a variable
That's not true. The variable is assigned a defined value because 5 > 6 evaluated to a defined value. While Perl doesn't specify what false value it returns from it's operators, it usually returns the scalar sv_no, which is a dualvar that appears to be an empty string when treated as a string, and zero when treated as a number.
$ perl -wE'say "> ", "".( undef )'
Use of uninitialized value in concatenation (.) or string at -e line 1.
>
$ perl -wE'say "> ", "".( "" )'
>
$ perl -wE'say "> ", "".( 0 )'
> 0
$ perl -wE'say "> ", "".( 5>6 )'
> # Behaves as ""
$ perl -wE'say "> ", 0+( undef )'
Use of uninitialized value in addition (+) at -e line 1.
> 0
$ perl -wE'say "> ", 0+( "" )'
Argument "" isn't numeric in addition (+) at -e line 1.
> 0
$ perl -wE'say "> ", 0+( 0 )'
> 0
$ perl -wE'say "> ", 0+( 5>6 )'
> 0 # Behaves as 0

Answer 2

5 > 6 is not undefined, but it is a false value. In this case, a dualvar that acts like an empty string when used as a string, or 0 when used as a number.
Because the value is false, you can just do
if ( $wiz ) { ... }
If you really want $wiz to be undefined, you could do something like this.
my $wiz = 5 > 6 || undef;
Now, if the expression 5 > 6 is true, $wiz will be 1, otherwise it will be undef.

Answer 3

Boolean false and undef are two different things ...
comparison assignments are executed in Boolean context only returning boolean values only and undef is not boolean thing
chk this
if(defined false){
print "false is also defined thing.."
}else{
print "else .."
}
So back to question , seems like comparisons can't return undef