This is a precedence problem. See Symbolic Unary Operators in perlop.

When the expression
    \ $name || $default
is evaulated, the sub-expression  \ $name has highest precedence. A reference is always true, so the second sub-expression (i.e., the  $default scalar) is never evaluated.

I would prefer the approach given by moritz above, but if you absolutely must do code interpolation within a string, here are a couple of approaches: 

>perl -wMstrict -le
"my $default = q{''};
 for my $name ('', 'fred') {
   print qq{name = ${ \ do { $name || $default }}};
   print qq{name = @{[ $name || $default ]}};
   }
"
name = ''
name = ''
name = fred
name = fred

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In reply to Re: Interpolation requires a spurious variable. Why? by AnomalousMonk
in thread Interpolation requires a spurious variable. Why? by bradcan

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