Suppose we wanted to convert a mathematical expression like 3^4+(11-(3*2))/2 into a reverse polish notation expression to evaluate the answer. This is called an infix expression. To convert it(to be able to evaluate the expression as well), we will use shunting yard algorithm. This algorithm is stack based and also includes an output list. The stack will handle operators, and the output queue will handle numbers.

### Algorithm explained:

**Step 1:**

Scan the Infix expression from left to right for tokens (Operators, Operands & Parentheses) and perform the steps 2 to 5 for each token in the Expression

**Step 2:**

If token is operand, Append it in postfix expression

**Step 3:**

If token is a left parentheses “(“, push it in stack.

**Step 4:**

If token is an operator,

Pop all the operators which are of higher or equal precedence then the incoming token and append them (in the same order) to the output Expression.

After popping out all such operators, push the new token on stack.

**Step 5:**

If “)” right parentheses is found,

Pop all the operators from the Stack and append them to Output String, till you encounter the Opening Parenthesis “(“.

Pop the left parenthesis but don’t append it to the output string (Postfix notation does not have brackets).

**Step 6:**

When all tokens of Infix expression have been scanned. Pop all the elements from the stack and append them to the Output String.

The Output string is the Corresponding Postfix Notation.

**Step 7:**

Exit or evaluate rpn and return final result

**C++ implementation**

#include <iostream> #include <iterator> #include <stack> #include <sstream> #include <vector> #include <main.h> using namespace std; bool TryParse(const string &symbol); int Priority(const string &c); bool isOperator(const string &c); int main() { string infix = "3 ^ 4 + ( 11 - ( 3 * 2 ) ) / 2";//our infix expression istringstream iss(infix); vector<string> tokens;//store the tokens here while(iss) { string temp; iss >>temp; tokens.push_back(temp); } vector<string> outputList;//output vector stack<string> s;//main stack //operator: +, -, *, /, ^, () //operands: 1234567890 for(unsigned int i = 0; i < tokens.size(); i++) //read from right to left { if(TryParse(tokens[i])) { outputList.push_back(tokens[i]); } if(tokens[i] == "(") { s.push(tokens[i]); } if(tokens[i] == ")") { while(!s.empty() && s.top() != "(") { outputList.push_back(s.top()); s.pop(); } s.pop(); } if(isOperator(tokens[i]) == true) { while(!s.empty() && Priority(s.top()) >= Priority(tokens[i])) { outputList.push_back(s.top()); s.pop(); } s.push(tokens[i]); } } //pop any remaining operators from the stack and insert to outputlist while(!s.empty()) { outputList.push_back(s.top()); s.pop(); } for(unsigned int i = 0; i < outputList.size(); i++) { cout<<outputList[i]; } return 0; } bool TryParse(const string &symbol) { bool isNumber = false; for(unsigned int i = 0; i < symbol.size(); i++) { if(!isdigit(symbol[i])) { isNumber = false; } else { isNumber = true; } } return isNumber; } int Priority(const string &c) { if(c == "^") { return 3; } if(c == "*" || c == "/") { return 2; } if(c== "+" || c == "-") { return 1; } else { return 0; } } bool isOperator(const string &c) { return (c == "+" || c == "-" || c == "*" || c == "/" || c == "^"); }

This is great. but shunting yard method and your code creates postfix notation not prefix notation.

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Thanks for the feedback. I will make corrections to the title.

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Hello. The whole code is amazing, great job, but I’ve been wondering if you can replace typing the equation in program by typing it from the keyboard while running the program, so you can create something like a start for an advanced calculator. It turns out you can’t. I copied the entire code, but instead of line “string infix = “3 ^ 4 + ( 11 – ( 3 * 2 ) ) / 2”; i want to have string infix;

cin>>infix;

When I run the program, it gives me only the first token. Is there any chance of fixing that, or did I do something wrong? It would be really great if you would explain that to me

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It looks like you need to store that input somewhere than just using cin>>. Use an array or vector to store that input. Here’s a hint to what you’re looking for:

while (true)

{

cin >> number >> operator

vector.push(number)

stack.push(operator)

}

Hope this helps. Thanks for commenting!

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