... what if my is within the condition of a postfix if? ... is it valid?

This is the way I look at it: The thing to remember is that the modifier clause (if that's the proper terminology) of a modified statement modifies the behavior of the statement. In order to do so, the modifier clause must always be evaluated. The ambiguity in a statement like
    my $x if $some_conditional;
concerns whether (and when) the lexical definition is evaluated, but  $some_conditional (or whatever expression may be there) must always be evaluated.

In a statement like
    0 if my $x = 1;
there is no ambiguity: the lexical is always defined (and, in this example, initialized). Similarly, in the statement
    do_something($_) for my @ra = @rb;
the for-loop initialization expression  my @ra = @rb must always be evaluated (including the definition and initialization of its lexical) in order that for may be able to control (i.e., modify the behavior of) the statement.

So, is
    0 if my $x;
and its ilk valid? Unquestionably (and unambiguously and without deprecation) yes, I would say. (But I'm too lazy right now to dig up a documentation citation.)

Consider the following code. Also consider running | compiling it with  -MO=Deparse,-p added (often enlightening in these cases). 

c:\@Work\Perl\monks>perl -wMstrict -le
"no warnings 'syntax';
 ;;
 0 if my $x = 42;
 print qq{$x};
 ;;
 my @orig = qw(foo bar baz);
 tr{a-z}{A-Z} && printf qq{'$_' } for my @copy = @orig;
 print '';
 print qq{(@orig) (@copy)};
 ;;
 for (0 .. 2) {
   die 'Oops...' if my $x;
   print qq{\$x (still) undefined here} if not defined $x;
   $x = 42;
   print qq{\$x is $x here};
   }
"
42
'FOO' 'BAR' 'BAZ'
(foo bar baz) (FOO BAR BAZ)
$x (still) undefined here
$x is 42 here
$x (still) undefined here
$x is 42 here
$x (still) undefined here
$x is 42 here

[download]

(Also consider running your code with warnings enabled. And strict.)

In reply to Re^3: postfix syntax enlightenment by AnomalousMonk
in thread postfix syntax enlightenment by RobertCraven

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