Beefy Boxes and Bandwidth Generously Provided by pair Networks
There's more than one way to do things
 
PerlMonks  

Assignment to a value only if it is defined

by voj (Acolyte)
on Jan 20, 2010 at 10:34 UTC ( #818418=perlquestion: print w/ replies, xml ) Need Help??
voj has asked for the wisdom of the Perl Monks concerning the following question:

Hi! found myself coding a lot of ugly statements like this:

$foo = $bar if defined $bar;

This is fine unless $bar is a complex statement or a function that I don't want to call twice. Of course I could create a temporary variable which makes the code even uglier:

$tmp = bar(); $foo = $tmp if defined $tmp;

What I would like is a short and simple syntax where I only have to type each part of the conditional assignment once:

$foo =ifdef $bar

Any ideas? The code should be short and readable. Bonus points if $foo can also be a list.

Comment on Assignment to a value only if it is defined
Select or Download Code
Replies are listed 'Best First'.
Re: Assignment to a value only if it is defined
by jettero (Monsignor) on Jan 20, 2010 at 12:01 UTC
    If bar() is a complex function, then perhaps it should be the one returning the default value...

    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; }

    Or just do what the others said above (5.10+ only): $x = bar() // "default";

    -Paul

      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 $_;}
        Because the other part of the statement ($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};

        {local $_ = $bar; $foo = $_ if defined $_;}

        $foo = $_ for grep defined, expensive();
Re: Assignment to a value only if it is defined
by Corion (Pope) on Jan 20, 2010 at 10:40 UTC

    If you're using Perl 5.10, you have the defined-or operator //:

    $foo //= $bar; # equivalent to $foo = $bar if defined $bar;
      Wrong. That's equivalent to
      $foo = $bar unless defined $foo;
      What the OP wants is:
      $foo = $bar // $foo;
      which is equivalent to
      $foo = $bar if defined $bar;
      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).

Re: Assignment to a value only if it is defined
by ikegami (Pope) on Jan 20, 2010 at 19:01 UTC
    my $bar = expensive(); $foo = $bar if defined $bar;
    if (defined(my $bar = expensive())) { $foo = $bar; }
    $foo = $_ for grep defined, expensive();
    sub assign_ifdef { $_[0] = $_[1] if defined($_[1]) } assign_ifdef($foo, expensive());

    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;
    or
    if (defined(my $bar = expensive()) { $_ = $bar for $i1, $i2; }

    Update: Added second and third.

      Thanks! That perfectly fits to assign to a scalar:

      sub setifdef { $_[0] = $_[1] if defined($_[1]) } setifdef $foo, $bar;

      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;
Re: Assignment to a value only if it is defined
by girarde (Friar) on Jan 20, 2010 at 18:27 UTC
    Your temporary assignment approach looks pretty short to me, unless this is golf.

Log In?
Username:
Password:

What's my password?
Create A New User
Node Status?
node history
Node Type: perlquestion [id://818418]
Approved by marto
Front-paged by MadraghRua
help
Chatterbox?
and the web crawler heard nothing...

How do I use this? | Other CB clients
Other Users?
Others chilling in the Monastery: (6)
As of 2015-07-30 05:23 GMT
Sections?
Information?
Find Nodes?
Leftovers?
    Voting Booth?

    The top three priorities of my open tasks are (in descending order of likelihood to be worked on) ...









    Results (270 votes), past polls