[ruby.parslet] Re: How to parse expression in Fortran?

[Kaspar Schiess]

I just dabbled with the bug in bracketed:

     def bracketed(thing, tag = false, left_bracket = '(', right_bracket 
= ')')
         thing = str(thing) if thing.kind_of? String
         left = spaced(left_bracket)
         right = spaced(right_bracket)

         if tag
             return (left >> thing >> right) | thing
         else
             return (left >> thing >> right)
         end
     end

The above will work a lot better and solve a few of the edge cases 
you've been having.

regards,
kaspar