Hi all,
Kindly go through the below codes:
use warnings;
use strict;
my $string = "test";
if ($string eq "test")
{
print "correct";
}
Output:
Correct
Now when I write the same if condition in program as below, I get warning
along with output.
use warnings;
use strict;
my $string = "test";
$string eq "test" ? print "correct" : "";
Output:
Correct
Useless use of constant in void context at line 5.
Can any one suggest the reason of warning in Case2.
Thanks & Regards,
Sanket Vaidya
http://www.patni.com
World-Wide Partnerships. World-Class Solutions.
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