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Re: defining variables using my - subroutines

by daxim (Chaplain)
on Sep 05, 2012 at 13:51 UTC ( #991862=note: print w/replies, xml ) Need Help??

in reply to defining variables using my - subroutines

The question is nonsensical because subroutines in Perl only take and return lists, not variables. If you define tightly-scoped variables in a subroutine, they are not available outside of it. Instead of paraphrasing your approaches in English prose, show the code/attempts you already have, then we can properly discuss it.
  • Comment on Re: defining variables using my - subroutines

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Re^2: defining variables using my - subroutines
by Freezer (Sexton) on Sep 05, 2012 at 14:13 UTC
    Surely they take variables. I don't understand when you say that they don't. If I define variables outside of a subroutine, are they available inside the subroutine?
      Under normal usage, they take values not variables. If you run the code
      sub add_one { my $x = shift; $x += 1; return $x; }
      You would not expect the variable you passed in (my $y = add_one($outer_x);) to change value, right? This is a little awkward/contrived, but this is in contrast to more object-oriented frameworks.

      Advanced: Of course Perl actually does pass by reference with lvalues bound to @_. But that's probably the best reason for transferring subroutine inputs to locally-scoped variables.

      #11929 First ask yourself `How would I do this without a computer?' Then have the computer do it the same way.

      $scalar = ( 1, 2, 3 );
      $scalar is a variable, it exists past this line, but ( 1, 2, 3 ); is a list of value, it does not exist past that line

      @array = ( 'list item 1', 2, 'another value', 'yet another value ');
      an @array is a variable, it has a name, it exists past the line, it is a bucket, you store scalars inside

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