True. It gets ambiguous when doing n*(-(x + y)) i.e. n x y + - *
(fail). The simplest solution is n 0 x y + - *
I can't actually think of any other unary operators.
On Fri, Apr 2, 2021 at 10:54 AM Richard Damon
One problem with trying to mix RPN with in-fix is that some operators, like - can be either a unary or binary operation in the in-fix notations, distinguished by context. In RPN you have lost that context.
is x y - - the same as -(x-y) or is it x-(-y) ? or is it waiting for another operator and be x ?? (-(-y)) or is it expecting a previous operand and is ?? - (x-y)
(same with +)
Typical RPN systems get around this by having unary - be a different symbol/key that binary -
-- Richard Damon
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