RPN Square Root

[Paultaker]

Hello guys, I'm studying java and I recently discovered the RPN, I was wondering how do add a square root in an RPN?


EDIT: Figured it out XD hehe

EDIT: I THOUGHT I FIGURED IT OUT help

Spoiler: CODE:

public static double SolvePostfix(String[] tokens) {
        double returnValue = 0;
        String operators = "+-*/^%";

        Stack<String> stack = new Stack<String>();

        for (String t : tokens) {
            if (!operators.contains(t)) {
                stack.push(t);
            } else {double a = Double.valueOf(stack.pop());
               
                               
                                    double b = Double.valueOf(stack.pop());
                                    switch (t) {
                                        case "+":
                                                stack.push(String.valueOf(a + b));
                                                break;
                                        case "-":
                                                stack.push(String.valueOf(b - a));
                                                break;
                                        case "*":
                                                stack.push(String.valueOf(a * b));
                                                break;
                                        case "/":
                                                stack.push(String.valueOf(b / a));
                                                break;
                                        case "^":
                                                stack.push(String.valueOf(pow(a,b)));
                                                break;
                                        case "%":
                                                stack.push(String.valueOf(b%a));
                                                break;
                                    }
                           
            }
        }

How do I evaluate unary operators?

 

[Aniki]

Square root and unary minus would simply pop only 1 argument from the stack and push their result (1 argument function = unary function, 2 argument function = binary function, I think?).

 

[GhostWolf]

You need a way to differentiate between a normal minus and an unary one. Assuming your tokenizer knows the difference, simply use your own imaginary operator, which the solver then knows to use, by popping only one argument.

I'd do it by having an additional list for all unary operators (which, again, are any character you want), and after you get the first argument, check if the operator is unary or not, and respond appropriately.

 

[Paultaker]

That's why I did, If token == 'symbol here' then ( a = pop then return a(sqrt) ) but it gives me an empty stack exception I don't know what I did wrong

EDIT: lets say the new symbol is x, infix: x ( 9 * 6), the postfix looks like this 6 9 * x when I print it. Is this correct?

 

[Dr Super Good]

You could translate square root into a ^ operation by inserting a 0.5 constant eg "9 0.5 sqrt". Likewise unary minus could be implemented by translation "0 9 -"

 

[Paultaker]

You could translate square root into a ^ operation by inserting a 0.5 constant.

Ohhh, I'll try that when I get home thanks, but I wanna learn how to make unary operators work in general too

 

[Dr Super Good]

but I wanna learn how to make unary operators work in general too

You need to make the operator pop the stack, and not automatically pop the stack.

 

[Paultaker]

What? What do you mean by automatically pop the stack and operator popping the stack?

 

[Dr Super Good]

The way your code looks at the moment it pops the arguments from the stack and then resolves which operation to run. To support unary operators in an efficient way you need to resolve which command to run and then pop the arguments it requires.

 

[Paultaker]

I figured out why it doesn't work. It seems that the square root symbol can't be read, I replaced it with s and now it works.

EDIT: Now my problem is, What precedence and associativity will I give to the unary operators

EDIT 2: How can I make: x ( ( 1 - c ( 30 ) ) / 2 ) to postfix: 1 30 c - 2 / x, it gives me this postfix: 1 30 2 / c - x