use strict; use warnings;

if('test' eq 0){  print "in if"   }
else           {  print "in else" }

if(!'test')    {  print "again in if"   }
else           {  print "again in else" }

__END__

prints: in else again in else

[download]

All strings in perl are evaluated to true (even the string 'false'!) except this: '0'

if('0' eq 0)  {  print "in if"   }
else          {  print "in else" }

__END__

prints: in if

[download]

the problem here is that you are trying to mix a boolean and a number, this is a logical operation, the same as if(TRUE is FALSE)... not numerical as in if (2*4 == 8), you are trying to say: if (TRUE == 8)

If you really want to specify that 'this_chain' == zero (the number) and force a numeric comparison you need to comment and deactivate use warnings, (but this is nonsense, use a variable name instead)

Updated: if ($var == 0) is not a problem for warnings, of course

In reply to Re: Why is this evaluating to true? by pvaldes
in thread Why is this evaluating to true? by pmarcoen

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