If that's not reasonable, then perhaps you need a default wrapper or something... Here's two ideas:

sub mybar {
   my $t = bar();
   return "default" unless defined $t;
   return $t;
}

sub default {
    my $to_call = shift;
    my $default = shift;
    my $t = to_call->(@_);
    return $default unless defined $t;
    return $t;
}
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Or just do what the others said above (5.10+ only): $x = bar() // "default";

The "default" action is not to do a assignement at all. But using the current value of $foo will do it as JavaFan suggested:

$foo = $bar // $foo;

However this is not much better then

$foo = $bar if defined $bar;

Because the other part of the statement ($foo) is duplicated - if I have a long variable like $myhash{mykey}->{mykey2} instead of $foo it only gets uglier. I was looking for a simple and short syntax like:

$foo =// $bar

But appereantly such operator does not exist in Perl. The best solution I found so far (also ugly but not getting more complex if $foo and $bar are long expressions) is:

{local $_ = $bar; $foo = $_ if defined $_;}
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Because the other part of the statement ($foo) is duplicated - if I have a long variable like $myhash{mykey}->{mykey2} instead of $foo it only gets uglier

Then try:

$_ = $bar // $_  for $myhash->{mykey}{mykey2};
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{local $_ = $bar; $foo = $_ if defined $_;}

$foo = $_ for grep defined, expensive();
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If you're using Perl 5.10, you have the defined-or operator //:

$foo //= $bar; # equivalent to $foo = $bar if defined $bar;
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Wrong. That's equivalent to

    $foo = $bar unless defined $foo;
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What the OP wants is:

    $foo = $bar // $foo;
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which is equivalent to

    $foo = $bar if defined $bar;
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assuming the absence of ties and overloading.

not quite — $foo //= $bar would test if $foo is defined, not $bar (it's equivalent to $foo = $bar unless defined $foo).

my $bar = expensive();
$foo = $bar if defined $bar;
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if (defined(my $bar = expensive())) {
   $foo = $bar;
}
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$foo = $_ for grep defined, expensive();
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sub assign_ifdef { $_[0] = $_[1] if defined($_[1]) }

assign_ifdef($foo, expensive());
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The preceeding solution is very similar to what you asked (=ifdef) and very simple (single op).

Bonus points if $foo can also be a list.

I'm not sure if you mean

my $bar = expensive();
($i1, $i2) = $bar if defined $bar;
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if (defined(my $bar = expensive()) {
   $_ = $bar for $i1, $i2;
}
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Update: Added second and third.

Thanks! That perfectly fits to assign to a scalar:

sub setifdef { $_[0] = $_[1] if defined($_[1]) }
setifdef $foo, $bar;
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And in case of a list I'll use the (slightly less readable) 'for grep defined,' psuedo-inline operator. It only makes sense with list references (if you treat (undef) as valid list). Defined scalar values which are transformed to a list can also be handled:

@list = @{$_} for grep defined, $listref_or_undef;
@list = split(",",$_) for grep defined, $splitme_if_defined;
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