Why do the '<' and 'lt' operators return different results in Perl?

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.

Answer 2

Since Perl is loosely typed, and values can silently convert between strings and integers at any moment, Perl needs two different types of comparison operators to distinguish between integer comparison (<) and string comparison (lt). If you only had one operator, how would you tell the difference?

Answer 3

Does this mean the 'lt' operator always converts the variables to string and then compare?

Yes, see perlop

What is the rationale for Perl making 'lt' operator behave differently from '<' operator?

Because having a numeric comparison operator and a string comparison operator makes a lot more sense then having a mumble mumble operator and another, identical mumble mumble operator.

Answer 4

Rationale? It's a string operator. From "perldoc perlop":

Binary "lt" returns true if the left argument is stringwise less than the right argument.

If that's not what you want, don't use it.

Answer 5

lt compares values lexically (i.e. in ASCII/UNICODE or locale order) and < compares values numerically. Perl has both operators for the same reason "10" + 5 is 15 rather than a type error: it is weakly typed. You must always tell the computer something unambiguous. Languages that are strongly typed tend to use casting to resolve ambiguity, whereas, weakly typed languages tend to use lots of operators. The Python (a strongly typed language) equivalent to "10" + 5 is float("10") + 5.