[perl #42300] [PATCH] t/pmc/sub.t: test for creation of lex by clone op
On Sat Sep 08 07:07:13 2007, bernhard wrote: As the original issue has been clarified, can I close this ticket? Since no one has suggested otherwise since September, I'll close it for you :-) kid51
[perl #42300] [PATCH] t/pmc/sub.t: test for creation of lex by clone op
On Do. 26. Apr. 2007, 16:45:08, [EMAIL PROTECTED] wrote: Now it makes sense. :) Anyway, I found this by following the Compiler FAQ, which says that a new closure should be created by cloning the sub. I think it should be changed to say newclosure, or even explain this (because you might really want to clone the Sub in some cases.) Hi, in r21136 I have changed compiler_faq.pod, so that it now uses 'newclosure' instead of 'clone'. More explanation could also be added to 'examples/tutorial/80_closure.pir'. As the original issue has been clarified, can I close this ticket? Regards, Bernhard -- /* [EMAIL PROTECTED] */
Re: [perl #42300] [PATCH] t/pmc/sub.t: test for creation of lex by clone op
On 4/26/07, Matt Diephouse via RT [EMAIL PROTECTED] wrote: First, the test (rearranged to include only the relevant parts): +.sub main :main +.local string ok, not_ok +ok = ok +not_ok = not ok + +# if 'not ok' is printed, it means that the lexical environment +# for the first closure in each pair, (where out = ok) +# was overwritten by the lexical environment created for the +# second closure (where out = not ok) + +$P10 = makebar_clone(ok) +$P20 = makebar_clone(not_ok) +$P10() +.end + +.sub makebar_clone +.param pmc out +.lex 'out', out +.const .Sub s = 'bar' +$P0 = clone s +.return($P0) +.end + +.sub bar :outer(makebar_clone) +$P0 = find_lex 'out' +say $P0 +.end (This prints not ok. The test in the patch expects ok.) You're arguing that the different copies of bar that are returned from makebar_clone should have different lexical environments. I'm pretty sure that this is not the case. Without using newclosure, there's no closure so the lexical environments are the same. What the :outer does in this case is rearrange the lexical stack so that makebar_clone appears in the lexical stack for bar. So we're using the lexical environment from the last time that makebar_clone was called. It's bizarre that this even works because without the closure, I'd think that the lexical environment would have destroyed. I'm not sure how intentional this is. The PDD isn't clear (to me) about what :outer means in the absence of newclosure. I'd definitely be interested in seeing why this would be a useful feature. More detail in the PDD would be nice. Thanks for the interesting patch. -- Matt Diephouse Now it makes sense. :) Anyway, I found this by following the Compiler FAQ, which says that a new closure should be created by cloning the sub. I think it should be changed to say newclosure, or even explain this (because you might really want to clone the Sub in some cases.)
[perl #42300] [PATCH] t/pmc/sub.t: test for creation of lex by clone op
First, the test (rearranged to include only the relevant parts): +.sub main :main +.local string ok, not_ok +ok = ok +not_ok = not ok + +# if 'not ok' is printed, it means that the lexical environment +# for the first closure in each pair, (where out = ok) +# was overwritten by the lexical environment created for the +# second closure (where out = not ok) + +$P10 = makebar_clone(ok) +$P20 = makebar_clone(not_ok) +$P10() +.end + +.sub makebar_clone +.param pmc out +.lex 'out', out +.const .Sub s = 'bar' +$P0 = clone s +.return($P0) +.end + +.sub bar :outer(makebar_clone) +$P0 = find_lex 'out' +say $P0 +.end (This prints not ok. The test in the patch expects ok.) You're arguing that the different copies of bar that are returned from makebar_clone should have different lexical environments. I'm pretty sure that this is not the case. Without using newclosure, there's no closure so the lexical environments are the same. What the :outer does in this case is rearrange the lexical stack so that makebar_clone appears in the lexical stack for bar. So we're using the lexical environment from the last time that makebar_clone was called. It's bizarre that this even works because without the closure, I'd think that the lexical environment would have destroyed. I'm not sure how intentional this is. The PDD isn't clear (to me) about what :outer means in the absence of newclosure. I'd definitely be interested in seeing why this would be a useful feature. More detail in the PDD would be nice. Thanks for the interesting patch. -- Matt Diephouse
[perl #42300] [PATCH] t/pmc/sub.t: test for creation of lex by clone op
# New Ticket Created by Yehoshua Sapir # Please include the string: [perl #42300] # in the subject line of all future correspondence about this issue. # URL: http://rt.perl.org/rt3/Ticket/Display.html?id=42300 --- t/pmc/sub.t 2007-04-04 17:20:12.0 +0300 +++ t/pmc/sub.new.t 2007-04-04 17:22:18.0 +0300 @@ -7,7 +7,7 @@ use lib qw( . lib ../lib ../../lib ); use Test::More; -use Parrot::Test tests = 61; +use Parrot::Test tests = 62; use Parrot::Config; =head1 NAME @@ -1379,6 +1379,57 @@ parrot;Foo;Bar OUTPUT +pir_output_is( 'CODE', 'OUTPUT', 'newclosure/clone Sub lex creation' ); +.sub main :main +$P1 = new .String +$P1 = 'ok' +$P2 = new .String +$P2 = 'not ok' + +# if 'not ok' is printed, it means that the lexical environment +# for the first closure in each pair, (where out = ok) +# was overwritten by the lexical environment created for the +# second closure (where out = not ok) + +$P10 = makefoo_newclosure($P1) +$P20 = makefoo_newclosure($P2) +$P10() + +$P10 = makebar_clone($P1) +$P20 = makebar_clone($P2) +$P10() +.end + +.sub makefoo_newclosure +.param pmc out +.lex 'out', out +.const .Sub s = 'foo' +$P0 = newclosure s +.return($P0) +.end + +.sub makebar_clone +.param pmc out +.lex 'out', out +.const .Sub s = 'foo' +$P0 = clone s +.return($P0) +.end + +.sub foo :outer(makefoo_newclosure) +$P0 = find_lex 'out' +say $P0 +.end + +.sub bar :outer(makebar_clone) +$P0 = find_lex 'out' +say $P0 +.end +CODE +ok +ok +OUTPUT + # Local Variables: # mode: cperl # cperl-indent-level: 4
