Here is how I think about for() assignments:

For a code like,

    for ($passed_in)
    {
        # inside BLOCK
    }
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Inside the BLOCK, $_ is equal to $$ref where $ref = \$passed_in
Except that if you did change $_, you will also modify $passed_in. They are magically tied together, $_ and the reference, that is.

It is faster this way, since only a REFerence is passed_in. No copy of $passed_in is made. However, for the following code,

    for my $scoped_copy ($passed_in)
    {
        # inside BLOCK
    }
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A copy of $passed_in is made to $scoped_copy.

The same reasoning above applies to other loop constructs, sub arguments, or CODE BLOCKS of map(), grep(), sort(), etc. for the $_ @_ $a $b variables.

Again, this is just my observation. Would someone with knowledge of Perl guts confirm that I could continue with this train of thought?