Re: Missing ref() return possibility?

in reply to 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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