Re^3: How do I pretend a reference isn't a reference

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Re^4: How do I pretend a reference isn't a reference by clinton (Priest) on Nov 06, 2008 at 12:54 UTC
It depends what `sub _` is defined to be, at compile time :) Check my code in Re^2: How do I pretend a reference isn't a reference and you'll see that `$foo` WOULD contain an object there.	[reply] [d/l] [select]
Re^5: How do I pretend a reference isn't a reference by clinton (Priest) on Nov 06, 2008 at 14:37 UTC
That's not what my code does, nor was I saying that $c was one thing at compile time and one at run time. This is what I do: `# init the app sub _ { 'compile time '.@_}; my $compile = _(1); sub foo { _(1) }; # finished compiling, so redefine sub _ sub _ { 'run time '.@_}; # run the code my $run = foo(); print " $compile - $run\n" __END__ compile time 1 - run time 1` [download] UPDATE: Meant to reply to Re^5: How do I pretend a reference isn't a reference	[reply] [d/l]
Re^5: How do I pretend a reference isn't a reference by Anonymous Monk on Nov 06, 2008 at 14:00 UTC
No it doesnt. `sub _ { 'compile time '.@_}; local *_ = sub { 'run time '.@_}; # the above is what your module/init code does my $c = _(1); die $c; __END__ run time 1 at - line 6.` [download] You were saying that $c is one thing at runtime, another at compile time. Its only something at runtime.	[reply] [d/l]