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

I have never seen this before: 
$uptime=`cat /proc/uptime`;
print ref(\$uptime)," $uptime\n";
$uptime*=1;
print ref(\$uptime)," $uptime\n";

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which prints 
SCALAR 12844.72 24452.05

SCALAR 12844.72

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It's not really a dereference is it? This is perl 5.10-comes-with-CentOS, so the * dereference operator had not been invented yet. Or is this a variation on *foo{THING}? Even more odd: 
$uptime=`cat /proc/uptime`;
print ref(\$uptime)," $uptime\n";
$uptime*=2;
print ref(\$uptime)," $uptime\n";

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Returns 
SCALAR 12956.65 24667.72

SCALAR 25913.3

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No idea where that second value of $uptime comes from ... I would do this with a regular expression myself, much more obvious ... Thanks in advance, w

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Re: What is the * postfix operator, pre-'feature postderef'?
by roboticus (Chancellor) on Sep 10, 2014 at 20:40 UTC

    wsanders:

    No, it's a shortcut. It turns out that:

    $uptime *= 2;

    is a shortcut for:

    $uptime = $uptime * 2;

    There's also +=, etc. See perldoc perlop for all the details.

    ...roboticus

    When your only tool is a hammer, all problems look like your thumb.

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      Gaaak I feel STUPID! Of course you are correct. The lack of spaces threw off my disused spatial recognition skills. And I just got new glasses, sheesh .... But it also threw me off that multiplying a string containing spaces by 1 results in the first "field" of the string, the same as 's/\s.*//'. One really ought to use: 
      $uptime=`cat /proc/uptime`;
print ref(\$uptime)," $uptime\n";
$uptime =~ s/\s.*//;
print ref(\$uptime)," $uptime\n";

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        If you want the first numeric field I'd use:

        ($uptime) = $uptime =~ /(\d+)/;

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        which makes the intent much clearer. What's with the ref(\$uptime)?

        Perl is the programming world's equivalent of English
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