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

[download]

In reply to Re^6: Perl can't make some easy arithmetics :( by no_slogan
in thread Perl can't make some easy arithmetics :( by rsFalse

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