Howdy, What does
foo [1] parse as? foo could be a variable and then foo is a simple array var: foo = [a,b] foo[1] If foo is a method, it could return an array and then you could access the 1st element: def foo return [a,b] end foo[1] It could also be a method with args: def foo x ... end foo [1] # same as foo([1]) This 5x ways to do anything syntax is truly a burden for the implementor (and as a user one of the reasons I hated perl). Regardless, is this a true *syntactic* ambiguity as I suspect? I.e., you need to simply pick an interpretation by dictum in the manual? Thanks, Ter _______________________________________________ Rubygrammar-grammarians mailing list [email protected] http://rubyforge.org/mailman/listinfo/rubygrammar-grammarians
