Now, what I don't get is, if within the function, I declare the $val as my $val (lexical scope), no warning message is shown. Nothing like "masking earlier declaration of $val". It just treats it as if it were a different variable.

see Coping with Scoping for a much better explanation than I could ever give. But I'll try, so that the explanations stay specific to what you're asking about.

When you declare the $val as lexical scope, its lexical scope is limited to the tightest enclosing block (in this instance, the file, or the first or second functions). A deeper scope (for example, a function inside the file-scope, or a block inside the function-scope) will inherit the value from its containing scope, unless it has a lexically-scoped variable of the same name. In the first program, your only scope for $var was the file-level scope (0x1c307c8), so when you edit $var inside the functions, it's editing the file-scope variable, even from the functions. But in the second program, the first subroutine has its own lexical $val at 0xb2d888, so anything you do to that is limited to that scope; similarly, second's 0xb2e290 is limited to its own scope.

This example shows some of the scopes, and when a "masks earlier declaration" will arise:

use strict; use warnings;
my $val = 1;
print "my own scope: $val\n";
{
  my $val = 2;
  print "my own scope: $val\n";
  {
    print "inherit: $val\n";        # this is another scope, but doesn
+'t have an override of the val=2
    $val = 2.5;
    print "edit:    $val\n";        # now it's a different value
  }
  print "keep the value edited: $val\n";    # which it keeps here
  {
    my $val = 3;                    # this is another scope with it's 
+own lexical $val, so the previous $val's are invisible to me
    print "my own scope: $val\n";
  }
  my $val = -2;     # warning: masks earlier declaration in same scope
  print "when scopes collide: $val\n";
}
print "back in original file scope: $val\n";
[download]