Re^6: Variable assignment error checking

[hdb]

It is not! In this expression:

$var = func($var) || $var;

the return value of func($var) will be assigned to $var if func($var) is true, otherwise the value of $var. In this expression

$var ||= func( $var );

the return value of func($var) will be assigned to $var if $var is false.

Completely different logic!

The expression $var ||= func( $var ); is shorthand for $var = $var || func( $var );. Can you spot the difference?

[Anonymous Monk]

$var = $var || func( $var );
$var = func( $var ) || $var;

They're certainly logically equivalent if we disregard the short-circuiting nature of operators. (Though the mathematics branch of logic generally deals with no other return values than true and false...)

[hdb]

As you say, only if we only worry about true or false...

my $var = 5;
sub func { 10 };

print $var || func($var), "\n";
print func($var) || $var, "\n";