That's were I was going (where I wanted to end up) as I was trying to remove the duplication in the code. The 2 mentions of the L-value "$pkg->{DIST} were definite candidates for reduction, but I got stuck along the way when the less 'reduced' expression I initially posted wasn't working as expected...Perhaps someone can point out the "logical error" from the following change of "logic" (I'm paying careful attention to op-precedence as others have suggested that might be the problem). BTW, "?:" has higher precedence than assignment "=". "and" is lower than both. From the orig (dropping unnecessary parens around assign before 'and' (numbering for greater easy in referring back to which step is causing problem) (0): 
return ( $rh=$pkg->{'RPM_HDR'} and $dist=$rh->tag('DISTRIBUTION') )
    ?    $pkg->{'DIST'}=$dist
    :    $pkg->{'DIST'}=$undef;

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Logically substituting "X" for the conditional expression in parens, "( $rh=$pkg->{'RPM_HDR'} and $dist=$rh->tag('DISTRIBUTION') )", I get (1): 
return X ? $pkg->{'DIST'}=$dist : $pkg->{'DIST'}=$undef;

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AHAH!....I see the prob...that "=" on the "end" (before $undef)... It is lower precedence than the ?: (as mentioned above...but wasn't sure which '=' the '?:' was interfering with). So the above (using D for "$pkg->{'DIST'}) is really (2?): 
return  (X ? D=$dist : D )  = $undef;

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Um...Why wouldn't that be a syntax error? the "(X?D=$dist:D)" part isn't a valid address that can be assigned to. (Note. "$undef" != undef, it's equal to a string of length 5 characters, 'undef'). I can see why what I wrote would be wrong -- but why wouldn't Perl flag it as illegal syntax?

In reply to Re^2: faulty expression prob by perl-diddler
in thread faulty expression prob by perl-diddler

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