in reply to Re: Scope, package, and 'my' variables
in thread Scope, package, and 'my' variables

Okay, so without the extra braces, each my $var gets tripped up by the previous definition.

But with the bare blocks, why does the aaa section associate and print a value for $var while in the bbb section this fails? The code appears identical and independent to me.

Update: That is, by putting braces around the section in package bbb that newly defines(?) a new lexical variable that also happens to be called $var, why doesn't this get set up independently from the $var of package aaa?

package bbb;
{
  use strict;
  use warnings 'all';
  my $var = 'second time for var';

  sub do_b {
    print "In 'bbb' var is set to '$var'\n";
  }
}

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Further update: I suppose it's bad form to do this via an Update, but the following notes don't stand out and they contain the critical pieces (at least for me.)

chromatic writes: It does declare a new lexical $var over which do_b closes. However, you call the subroutine before the assignment occurs.

Errto writes: The issue is that the assignment of $var within the block under bbb never happens before you call do_b. Don't think of a line like my $x = somevalue; as a declaration of a variable to be initialized before the program begins. Think of it as a statement that runs when the program gets to it.

Certainly these "side effects" would not happen if I structured the components of the example more logically, i.e. saved that "main" section until the end. But it's exactly that jumble that leads to the strange (for me) behavior that now lets me grasp the concept. ++ to you and all who have contributed to this thread.

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Re^3: Scope, package, and 'my' variables
by friedo (Prior) on Jan 04, 2005 at 19:56 UTC
    This is due to closures. With the braces, your first sub, do_a is defined within the lexical scope that sets $var to 'first var'. In other words: 

    {  
  my $var = 'first var';  # sets $var for this block

  # defines a sub IN this block, so gets this $var
  sub do_a {
      print "In 'aaa' var is set to '$var'\n";
  }
}

    [download]

    Because subroutine names are available globally, do_a can be called from anywhere, but gets the $var defined in its lexical block.

    Similarly, do_b is defined in a block where $var is set to nothing, and do_c is in a block with no $var at all.

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[d/l]
      In my original question, yes, that makes sense. But in my clarifying question, in which there are braces around the code in package bbb, why doesn't that set up an independent version of $var which then should get picked up by do_b and thus print out "second time for var"?

      Thanks 

      package bbb;
{
  use strict;
  use warnings 'all';
  my $var = 'second time for var';

  sub do_b {
    print "In 'bbb' var is set to '$var'\n";
  }
}

      [download]
[reply]
[d/l]

        It does declare a new lexical $var over which do_b closes. However, you call the subroutine before the assignment occurs.

[reply]
[d/l]
[select]
        The issue is that the assignment of $var within the block under bbb never happens before you call do_b. Don't think of a line like my $x = somevalue; as a declaration of a variable to be initialized before the program begins. Think of it as a statement that runs when the program gets to it.
[reply]
[d/l]

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