Perl doesn't really make a distinction between "normal" blocks and subroutine blocks with regard to lexical scoping rules. You can even take and copy a reference to a subroutine that accesses lexicals in its outer scope that way. This is by design. See also
perlfaq7 (what's a closure?),
perlsub and
perlref.
use strict;
sub make_ref {
my $name = shift;
my $i = 99;
return sub {
print "ref $name: ",$i++,"\n";
}
}
my $ref1 = make_ref("sub1");
my $ref2 = make_ref("sub2");
for (1 .. 10) {
$ref1->();
$ref2->() if $_ % 2 == 0;
}
output:
ref sub1: 99
ref sub1: 100
ref sub2: 99
ref sub1: 101
ref sub1: 102
ref sub2: 100
ref sub1: 103
ref sub1: 104
ref sub2: 101
ref sub1: 105
ref sub1: 106
ref sub2: 102
ref sub1: 107
ref sub1: 108
ref sub2: 103
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