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

Is there any danger in doing something like this? 

$temp = defined($temp = &func()) ? $temp : 'undefined';

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When I say 'danger', I'm wondering if I may be surprised at my result due to the order of evaluation being different than I expect.

By the way, my motivation for this is to either save a little space by not declaring an extra variable: 

$temp1 = &func();
$temp2 = defined($temp1) ? $temp1 : 'undefined';

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or to save a little time by not calling func twice: 
$temp = defined(&func()) ? $func() : 'undefined';

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Replies are listed 'Best First'.
Re: Safe to use ternary operator like this?
by ChOas (Curate) on Nov 16, 2000 at 20:13 UTC
    I have to run, but I just saw this...
    this is how I would do it: 
    my $Temp=&Func() || 'undefined';

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    Bye!!
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      This will set $Temp to 'undefined' if &Func returns a false value (see What is true and false in Perl? for more details). I like the original code, methinks. It's just what the ternary operator was designed to do.

      Philosophy can be made out of anything. Or less -- Jerry A. Fodor

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      Not quite the same thing depending on the sub return though: 
      use strict;
my $temp;

$temp = defined ($temp = func()) ? $temp : 'undefined';
print "func:$temp\n";

$temp = func() || 'undefined';
print "func:$temp\n";

sub func { return ''; }

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      produces: 
      func:
func:undefined

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