in reply to Re^2: split and assign
in thread split and assign

Although it must be admitted that this approach would get a bit tedious if one were interested not in the second field but in the twenty-second!
I hoped for a mad moment that
my ( ( undef ) x 21, $var ) = func;
or at least
( ( undef ) x 21, my $var ) = func;
would work—but they don't ….