Hmm, thanks, very useful tip, but what about those operators at the end? Three plusses? What do they represent? Is it correct that

**z x * y + z x y + + +** parses into **(((z * x) + y) + z + x + y)**?

I have seen **x y * -** becoming **-x * y** in infix. Does that mean that operators at the end mean sign for individual variables, starting at the beginning? Or does that mean that you have to negate the result? :\

As far as your last question: no. That has to do with what the expression means.

Let's say you're parsing x y * -

You read x, and push it onto the stack. The stack is now x

You read y and push it onto the stack. The stack is now x|y

You read *. It's an operator. It's a binary operator so you pop two operands from the stack. The stack is now empty.

You calculate the result (x*y) and push it back in. The stack is now (x*y).

You read -. Let's say this symbol represents the unary operation, but not the binary (to avoid ambiguity). This is an unary operator, so you pop one element from the stack (and indeed that is all you can do. A binary operator at this point would have to be interpreted as a syntax error). The stack is now empty.

You calculate the result -(x*y) and push it back in. The stack is now -(x*y).

No more tokes on input, and one value on the stack means you've reached the end of a successful parse. You pop the final result: -(x*y)

Now note that -(x*y) is the same as -x*y = x*(-y) = x*y*(-1), etc, etc. This doesn't mean that a - at the end of the expression negates the first value. It just happens that, because of the way multiplication works, those two expressions are equivalent.