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( #3333=superdoc: print w/replies, xml ) Need Help??

Yes, there is a difference.

\() is like a factored out reference operator before every thing inside the parenthesis, so \($foo, @bar, %baz, &frah) is (\$foo, \@bar, \%baz, \&frah), except if there's only one aggregate data type inside it. In that case it gets flattened and the backslash is then applied to each element of that list, as you say. This means that

\((@foo), @bar)
(\(@foo), \@bar)
which becomes
(\($foo[0], $foo[1], ..., $foo[$#foo]), \@bar)
and finally
((\$foo[0], \$foo[1], ..., \$foo[$#foo]), \@bar)

Another example:

\((@foo, @bar), @baz) (\(@foo, @bar), \@baz) ((\@foo, \@bar)), \@baz)

Update: I just realized that my second example is rather useless as the end result is the same as if there was no inner parenthesis. But the intermediate step perhaps still has a pedagogical value.

(I don't know if this recursive behaviour is how it's actually performed under the hood, but it works this way none the less.)


In reply to Re^2: Your favourite gory detail... by ihb
in thread Your favourite gory detail... by Ido

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