the last case is the interesting one for me, the coderef of a block at a fixed position (and therefore the same code). I don't care if other blocks get the same ref after this one gets destroyed.

And my test with your code shows that the ref is fix if its not returned:

  DB<123> sub test(&) { my( $c ) = @_; print 0+$c,$/; return $c }

  DB<124> ;{ my @a; for(1..3){ my $a; push @a, test {$a} } }
145187472
145182504
144725160
 => ""

  DB<125> sub test(&) { my( $c ) = @_; print 0+$c,$/}

  DB<126> ;{ my @a; for(1..3){ my $a; push @a, test {$a} } }
145184216
145184216
145184216
 => ""
[download]

So the ref is chosen dynamically from a pool of available refs, it can be the last one if it has been released before.

Thx! =)

UPDATE

OK the following code shatters all my hopes that I found an answer to an old question ...

  DB<155> sub tst2(&) { my( $c ) = @_; print &$c,":\t",0+$c,$/;return 
+$c}

  DB<156> sub tst1(&) { my( $c ) = @_; print &$c,":\t",0+$c,$/;}

  DB<157> ;{ my $b; for my $a (1..3){tst1 {$a} ; $b=tst2 {42} if $a==1
+ } }
1:    145234328
42:    145234328
2:    141318480
3:    141318480
[download]

it doesn't matter how tst1() deals with the coderef, any later coderef generation might bind the address and disable it for other use.

> If a sub doesn't close over any lexicals, than I bet that refs to it will be constant.

Nope it doesn't matter if there are closed over variables, unfortunately you loose your bet...

  DB<158> ;{ my $b; for my $a (1..3){tst1 {1} ; $b=tst2 {2} if $a==1 }
+ }
1:    141317840
2:    141317840
1:    141318608
1:    141318608
[download]

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