ref and qx

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

Hi to all !

when reading doc by "perldoc -f ref" I get, this sentence :

               The result "Regexp" indicates that the argument is a re
+gular
               expression resulting from "qr//".
[download]

So, for testing purpose, I tried this code :

#!/usr/bin/perl
my $regX = qx /\.pm/ ;
printf ("type of regX=%s\n", ref(\$regX));
[download]

And I get "type of regX=SCALAR" instead of the waited "Regexp".

What did I do wrong ??

Some infos:

Perl 5.12.3 from ActiveState

I tried {/\.pm/} with no more success.

Thanks to all !

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Re: ref and qx by ikegami (Patriarch) on May 06, 2011 at 05:40 UTC
First, you're using `qx` (aka `readpipe`) instead of `qr`. Second, `\$refX` is a reference to scalar `$refX`, and thus `ref(\$refX)` returns "`SCALAR`". If the value of `$refX` had been constructed using `qr`, `ref($refX)` would get you "`Regex`" since it contains a reference to a Regex. I recommend that you use `re::is_regexp` instead if you're using Perl 5.10+. This will catch both references to regex patterns and a regex patterns themselves. `>perl -E"$r=qr/a/; say ref($r) eq 'Regexp' \|\|0" 1 >perl -E"$r=qr/a/; say ref($$r) eq 'Regexp' \|\|0" 0 >perl -E"$r=qr/a/; say re::is_regexp($r) \|\|0" 1 >perl -E"$r=qr/a/; say re::is_regexp($$r) \|\|0" 1` [download]	[reply] [d/l] [select]
Re^2: ref and qx by Anonymous Monk on May 06, 2011 at 12:13 UTC
Shame on me !!!! And many thanks to you !	[reply]