Question

Please construct a code following the instructions below

Part 3 – infixEvaluator method

You are required to implement the following method:  

public static double infixEvaluator(String line)

This function first needs to tokenize the input expression (in the form of a String and stored in input variable “line”). You can tokenize the input expression by creating an object of StringSplitter and passing it the String as follows. StringSplitter uses a queue to accomplish the tokenization (see the class for details).

StringSplitter data = new StringSplitter(line);

Next, create your two stacks. Remember one will contain the operators and the other one will include the operands. Define them as follows:

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

Stack<Double> operands = new Stack<Double>();

Before we continue further in this method, it will be helpful to consider helper methods we might need (which will make our job easier :-).

Skeleton of the code below:

/**

* give a description of the purpose of this method

* @param line fill in

* @return fill in

*/

public static double infixEvaluator(String line){

return 0.0; // placeholder

}

Answer 1

Hello There,

Code for the above probem is given as :-

import java.util.Stack;

    public static double infixEvaluator(String line)

    {

        char[] tokenized = line.toCharArray();

         // to store operands..

        Stack<Double> operands = new Stack<Double>();

        // to store operators..

  Stack<Character> operators = new Stack<Character>();//taking of Character type as it is converted to character array above..

        for (int i = 0; i < tokenized.length; i++)

        {

             // if a current character is a whitespace just skip..

if (tokenized[i] == ' ')

                continue;

            // if a current character is a number just push it to stack of operators..

            if (tokenized[i] >= '0' && tokenized[i] <= '9')

            {

                StringBuffer S_B = new StringBuffer();

                // it is possible that it contains more than single digit numbers..

                while (i < tokenized.length && tokenized[i] >= '0' && tokenized[i] <= '9')

S_B.append(tokenized[i]);

i=i+1;

operators.push(Double.parseDouble(S_B.toString())); //converting each character to Double because calculations done in Double values..

            }

            // if an opening bracket is encountered push it to operands...

            else if (tokenized[i] == '(')

                operands.push(tokenized[i]);

            // For a closing bracket the whole brace is needed to be solved..

            else if (tokenized[i] == ')')

            {

                while (operands.peek() != '(')

operators.push(helper(operands.pop(), operators.pop(), operators.pop()));

                operands.pop();

            }

            // for operators we have : -

        else if (tokenized[i] == '+' || tokenized[i] == '-' || tokenized[i] == '*' || tokenized[i] == '/')

            {

                // Now until top of operands has same or greater precedence than current

                // character that is operator. Solve top two of operators as according to

                // top of operands ..

             while ( check_precedence(tokenized[i], operands.peek()) && !operands.empty() )

operators.push(helper(operands.pop(), operators.pop(), operators.pop()));

                // Push current character to operands...

                operands.push(tokenized[i]);

            }

        }

        // After parsing till here apply the available operands to remaining operators...

        while (!operands.empty())

operators.push(helper(operands.pop(), operators.pop(), operators.pop()));

// return the top of operators as an final answer...

        return operators.pop();

    }

    // this check_precedence function gives true if operand_2 has higher or same precedence as of operand_1 ,..

    // otherwise it simply returns false.

    public static boolean check_precedence(char op1, char op2)

    {  

  

if ((operand_1 == '/' || operand_1 == '*') && (operand_2 == '+' || operand_2 == '-'))

            return false;

         if (operand_2 == '(' || operand_2 == ')')

            return false;

        else

            return true;

    }

    // an helper function to calculate result of top two operators..

    public static Double helper(char operand, Double x, Double y)

    {

if(operand=='*')

return x * y;

else if(operand=='+')

            return x + y;

else if(operand=='-')

            return x - y;

else if(operand=='/')

{

  if (y == 0)

{

throw new

UnsupportedOperationException("divide by zero exception occur..");

}

            return a / b;

        }

        return 0.0; //placeholder

    }

This is the required code to solve the infix experession using two stacks...for any further doubt you can ask again...

Thanks for reaching out to us..

keep asking and keep learning... :)))))