in reply to Creating 'new' subroutine.

Is it just me or isn't that equivalent to: 
$options = { @_ };

[download]

Replies are listed 'Best First'.
Re^2: Creating 'new' subroutine.
by LanX (Saint) on Oct 11, 2013 at 10:21 UTC
    > Is it just me or isn't that equivalent to: $options = { @_ };

    Good point, w/o testing I'd say yes !

    Maybe the intention was to check if @_ has even number of elements?

    update

    Still with testing. 

      DB<112> use Data::Dump qw/pp/

  DB<113> sub tst { pp (scalar @_ ? {@_} : {}); pp {@_};()} 

  DB<114> tst
{}
{}

  DB<115> tst a
{ a => undef }
{ a => undef }

  DB<116> tst a,1
{ a => 1 }
{ a => 1 }

    [download]

    Cheers Rolf

    ( addicted to the Perl Programming Language)

[reply]
[d/l]
[select]

      If we're guessing at intentions, then I might wager:

      $options = scalar(@_) == 1 ? $_[0] : { @_ };

      [download]

      Though that's probably better written as:

      $options = { scalar(@_) == 1 ? %{$_[0]} : @_ };

      [download]

      ... to deliberately shallow clone in the case of a hashref.

      use Moops; class Cow :rw { has name => (default => 'Ermintrude') }; say Cow->new->name
[reply]
[d/l]
[select]
      Even if the intention was to write it with a check for evenness, @_ % 2 == 0 ? {@_} : {}, it would be very poor coding practise to just silently discard the options supplied by the user (if odd).
[reply]
[d/l]

      My own quick check confirms what LanX (Rolf) just posted. But with respect for checking for an even number of elements, I wouldn't bother.

      One would probably want the fatal error if an odd number was present sense one would most likely be expecting named args coming in. Letting it die is probably the correct course of action.

      The answer to the question "Can we do this?" is always an emphatic "Yes!" Just give me enough time and money.
[reply]
Re^2: Creating 'new' subroutine.
by Lennotoecom (Pilgrim) on Oct 11, 2013 at 10:03 UTC
    if the second element is empty
    then shift @_ moves to the $_ nothing
    which means scalar @_ returns zero
    in which case : {} adds into hash nothing,
    but creates reference to it
    so $options {@_} would do the trick, I guess

    please correct if it's wrong

    updated


    thanks for the correction snippet was incorrect
[reply]
      @_ is an array, an array in scalar context returns the number of elements (a number), and anumber is never a reference
[reply]
        yes but
        @a = ();
        $b = {@a};
        will create a reference to a hash
[reply]
      Thanks for the reply. I still am far from testing it as we are running on mac, and are struggling with the setup. So, do you mean scalar @_ determines how many elements to be added into hash while creating the reference?
[reply]

In Section Seekers of Perl Wisdom