Re: Missing ref() return possibility?

The types you list are the basics. All other cases are covered by "If the referenced object has been blessed into a package, then that package name is returned instead."

For regular expressions, this is demonstrated by:

use Scalar::Util qw/blessed/;
$q=qr//;
print ref($q);
print blessed($q);
__END__
Regexp
Regexp
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Comment on Re: Missing ref() return possibility? Download Code

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Re^2: Missing ref() return possibility? by kyle (Abbot) on May 16, 2008 at 21:51 UTC
What your code implies but isn't quite explicit about is that `qr()` actually gives back a blessed reference to a scalar. `use Scalar::Util qw/blessed reftype/; my $q = qr//; print 'ref: ', ref($q), "\n"; print 'blessed: ', blessed($q), "\n"; print 'reftype: ', reftype($q), "\n"; __END__ ref: Regexp blessed: Regexp reftype: SCALAR` [download] There's some more about this in How can I tell if an object is based on a regex?	[reply] [d/l] [select]

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Re^2: Missing ref() return possibility?
by kyle (Abbot) on May 16, 2008 at 21:51 UTC

What your code implies but isn't quite explicit about is that qr() actually gives back a blessed reference to a scalar.

use Scalar::Util qw/blessed reftype/;
my $q = qr//;
print 'ref:     ', ref($q),     "\n";
print 'blessed: ', blessed($q), "\n";
print 'reftype: ', reftype($q), "\n";
__END__
ref:     Regexp
blessed: Regexp
reftype: SCALAR
[download]

There's some more about this in How can I tell if an object is based on a regex?

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