No.

my $value = shift() || 'default';
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and

my $value = shift() or 'default';
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will not work the same.

my $value = shift() || 'default';
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will set $value to the value returned by shift() if it is true and to 'default' if it is false.

my $value = shift() or 'default';
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will set $value to the value returned by shift() no matter what!
Here is an example using a handy tool B::Deparse that will show you how the compiler read your program, the -p adds extra parentheses to make it very clear.

#temp.pl
$value = $x or $y or $z;
$value = $x || $y || $z;
$value = $x or $y || $z;
$value = $x || $y or $z;
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then run

perl -MO=Deparse,-p  temp.pl
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it outputs

((($value = $x) or $y) or $z);
($value = (($x || $y) || $z));
(($value = $x) or ($y or $z));
(($value = ($x || $y)) or $z);
temp.pl syntax OK
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if you notice

$value = $x or $y or $z;
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and

$value = $x or $y || $z;
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just set $value to the value of $x no matter what the value of $y or $z because or has a lower precidence than =.
Did you notice that

 
$value = $x || $y or $z;
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will just set $value to the value of $x || $y no matter what the value of $z?

--

flounder

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