Scalar alias vs. typeglob

borisz has asked for the wisdom of the Perl Monks concerning the following question:

Hi,
I just want to alias a scalar to a different package var. But only the * notation work as expected. This looks like a bug to me, but propably someone can explain the behavior.

I expect, that both aliases do the same and the output is 'boris boris'


package XYZ;
sub test { local *name = \'boris'; $_[1]->() }

package ABC;

*scalar_alias = \$XYZ::name;    # does not work
*glob_alias   = *XYZ::name;     # DWIM

package main;
use warnings;

XYZ->test(
  sub {
    print 'scalar_alias: ', $XYZ::name, ' ', $ABC::scalar_alias, $/;
    print 'glob_alias:   ', $XYZ::name, ' ', $ABC::glob_alias,   $/;
  }
);
__OUTPUT__
scalar_alias: boris 
glob_alias:   boris boris
Use of uninitialized value in print at untitled 74.pl line 15.
[download]

Boris

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Re: Scalar alias vs. typeglob by ambrus (Abbot) on Oct 05, 2006 at 11:54 UTC
The `XYZ::test` function has the statement `local name = \'boris';`. That statement chagnes the scalar reference part of `XYZ::name` not only the value of the scalar it points to. Now, the ABC package copies the old scalar reference to the *ABC::scalar_alias glob, so it won't see the new reference. (The same would be true if you changed that statement to `local $name = 'boris';` because the local keyword also replaces the scalar reference.)	[reply] [d/l] [select]

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Re: Scalar alias vs. typeglob
by ambrus (Abbot) on Oct 05, 2006 at 11:54 UTC

The XYZ::test function has the statement local *name = \'boris';. That statement chagnes the scalar reference part of *XYZ::name not only the value of the scalar it points to. Now, the ABC package copies the old scalar reference to the *ABC::scalar_alias glob, so it won't see the new reference. (The same would be true if you changed that statement to local $name = 'boris'; because the local keyword also replaces the scalar reference.)

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