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

In reply to Re: Using an "outer" lexical in a sub? by Joost
in thread Using an "outer" lexical in a sub? by cornballer

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