Yes, unless one of the variables has already been coerced to be a bigint, so even this is enough:

use bigint;
while(<>) {
    my ($a, $b)= split/ /;
    $a += 0;
    print $a - $b, "\n";
}
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However you can just coerce everything when you first read them in as strings:

use bigint;
while(<>) {
    my ($a, $b)= map { 0+$_ } split/ /;
    print $a - $b, "\n";
}
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To explain further: use bigint only causes numeric literals to become bigints. It does not change the meaning of operators (like -), no matter what the perldoc might lead you to believe. 0 is a literal, and becomes a bigint. $a and $b are strings, and do not become bigints unless you add 0 to them. Just saying "$a-$b" does not trigger bigint conversion.

Edit: Here's a quick demonstration:

> perl -le 'use bigint; print 22/7; print "22"/"7"'
3
3.14285714285714
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I have no time to look into the details but this looks like a flaw in the design.¹

According to it's POD is bigint overloading operators, so why restricting this to integers only?

Maybe it's because of a restriction in the overload mechanics but as I said I have no time to look into the details .... (and this place is full of honorable monks willing to answer it! ;)

Cheers Rolf

_{(addicted to the Perl Programming Language and ☆☆☆☆ :)}

update

¹) to elaborate further: Perl tries hard not to distinguish between different types of scalars, that's one of the reasons why it has a nummeric add + and a string concat . where languages like JS only have + for both.

But bigint seems to break this "paradigm".