Question

I am just learning Perl’s comparison operators. I tried the below code :-

$foo=291;
$bar=30;

if ($foo < $bar) { 
        print "$foo is less than $bar (first)\n"; 
}

if ($foo lt $bar) { 
        print "$foo is less than $bar (second)\n"; 
}

The output is 291 is less than 30 (second). Does this mean the lt operator always converts the variables to string and then compare? What is the rationale for Perl making lt operator behave differently from the < operator?

Thanks,

Answer 1

Your guess is right. The alphabetic operators like lt compare the variables as strings whereas the symbolic ones like < compare them as numbers. You can read the perlop man page for more details.

The rationale is that scalars in Perl are not typed, so without you telling it Perl would not know how to compare two variables. If it did guess then it would sometimes getting it wrong, which would lead to having to do things like ' ' + $a < ' ' + $b to force string comparsion which is probably worse than lt.

That said this is a horrible gotcha which probably catches out everyone new to Perl and still catches me out when coming back to Perl after some time using a less post-modern language.