The fact that f( ++$n, ++$n ) passes an alias to $n, rather than the value resulting from the preincrement, is just another broken behaviour.
Note that this behaviour isn't unique to ++. 
    sub f {say "@_"}
    my $n = 1;
    f $n += 2, $n += 2;
    __END__
    5 5

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In reply to Re^4: Pre vs Post Incrementing variables by JavaFan
in thread Pre vs Post Incrementing variables by SavannahLion

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