reverse polish notation fuction-CodePudding

a function evaluateCountdown that takes a string containing a mathematical expression written in Reverse Polish Notation, and returns the result of evaluating the expression, as a double.

In reverse polish notation (RPN), operators are written after rather than between what they act upon. For instance:

3.0 4.0

is the same as the usual (infix) notation (3.0 4.0). Or the below expression in RPN format:

3.0 4.0   2.0 *

is the same as (3.0 4.0) * 2

I am getting a logic error, the input "3.0 4.0 - 2.0 *" is supposed to give the output -2.0 however the output I am receiving when calling the main function is 0, the input string must be in the format where the numbers are doubles, the function works when the input string is, for example, "3 4 - 2 *" (returns a correct answer of -2), but not when in the form "3.0 4.0 - 2.0 *"

the function evaluateCountdown iterates through the input string and when it finds a number it pushes that number into the stack as a double, so for example when tokens[i] is 3 then 3 tokens[i 1] tokens[i 2], in this case '3' '.' '0' is pushed into the stack as a double. However, this isn't happening and I'm not sure why.

#include <string>
#include <sstream>
#include <vector>
#include <stack>
#include <iostream>
#include <iterator>
using std::vector;
using std::ostream;
using std::string;
using std::cout;
using std::stack;
using std::iterator;
using std::endl;
using std::stod;

double evaluateCountdown(string & tokens) {
    // this method does not work if the string is empty or a string of just spaces, string must be a expression in RPN
    stack<double> stack;
    for (int i = 0; i < tokens.size(); i  ) {
       
        if (tokens[i] == ' ')
        {
            double val1 =  stack.top();
            stack.pop();
            double val2 = stack.top();
            stack.pop();
            stack.push((val1   val2));
        }

        else if (tokens[i] == '-')
        {
            double val1 = stack.top();
            stack.pop();
            double val2 = stack.top();
            stack.pop();
            stack.push((val2 - val1));
        }

        else if (tokens[i] == '*')
        {
            double val1 = stack.top();
            stack.pop();
            double val2 = stack.top();
            stack.pop();
            stack.push(val1 * val2);
        }

        else if (tokens[i] == '/')
        {
            double val1 = stack.top();
            stack.pop();
            double val2 = stack.top();
            stack.pop();
            stack.push((val2 / val1));
        }

        else if (tokens[i] == ' ')
        {
            // if char is a space do nothing and move onto the next char.
        }

        else
        {
            string str = "";
            char c1, c2, c3;
            c1 = tokens[i];
            c2 = tokens[i   1];
            c3 = tokens[i   2];
            // concat chars togther to get required string eg '3'   '.'   '0'
            auto ans = string(1, c1)   c2   c3;
            str  = ans;
            double val = stod(str);
            stack.push(val);
        } 

        

        
    }
    double val = stack.top();
    stack.pop();
    return val;
   
}

int main() {
    string s = "3.0 4.0 - 2.0 *";
    double ans = evaluateCountdown(s);
    cout << ans << endl;
}

CodePudding user response：

When the code finds a number, it should increment i by two, unless doing that the code does not work.

Therefore, the else condition could be written as follows(The rest of the else statement will remain intact):

c1 = tokens[i];
c2 = tokens[  i];
c3 = tokens[  i];

If you are a new programmer, I would recommend that when you encounter a problem like that, you use a debugger to debug your program (even the most professional programmers need to debug their code).