Question

Write a program to convert the following infix to the postfix.

(A – B) * C + D / E * (F – G)

Using C++ will have to run on mac

Answer 1

#include<iostream>

#include<stack>

#include<string>

using namespace std;

// Function to convert Infix expression to postfix

string InfixToPostfix(string expression);

// Function to verify whether an operator has higher precedence over other

int HasHigherPrecedence(char operator1, char operator2);

// Function to verify whether a character is operator symbol or not.

bool IsOperator(char C);

// Function to verify whether a character is alphanumeric chanaracter (letter or numeric digit) or not.

bool IsOperand(char C);

int main()

{

string expression;

cout<<"Enter Infix Expression \n";

getline(cin,expression);

string postfix = InfixToPostfix(expression);

cout<<"Output = "<<postfix<<"\n";

}

// Function to evaluate Postfix expression and return output

string InfixToPostfix(string expression)

{

// Declaring a Stack from Standard template library in C++.

stack<char> S;

string postfix = ""; // Initialize postfix as empty string.

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

// Scanning each character from left.

// If character is a delimitter, move on.

if(expression[i] == ' ' || expression[i] == ',') continue;

// If character is operator, pop two elements from stack, perform operation and push the result back.

else if(IsOperator(expression[i]))

{

while(!S.empty() && S.top() != '(' && HasHigherPrecedence(S.top(),expression[i]))

{

postfix+= S.top();

S.pop();

}

S.push(expression[i]);

}

// Else if character is an operand

else if(IsOperand(expression[i]))

{

postfix +=expression[i];

}

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

{

S.push(expression[i]);

}

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

{

while(!S.empty() && S.top() != '(') {

postfix += S.top();

S.pop();

}

S.pop();

}

}

while(!S.empty()) {

postfix += S.top();

S.pop();

}

return postfix;

}

// Function to verify whether a character is english letter or numeric digit.

// We are assuming in this solution that operand will be a single character

bool IsOperand(char C)

{

if(C >= '0' && C <= '9') return true;

if(C >= 'a' && C <= 'z') return true;

if(C >= 'A' && C <= 'Z') return true;

return false;

}

// Function to verify whether a character is operator symbol or not.

bool IsOperator(char C)

{

if(C == '+' || C == '-' || C == '*' || C == '/' || C== '$')

return true;

return false;

}

// Function to verify whether an operator is right associative or not.

int IsRightAssociative(char op)

{

if(op == '$') return true;

return false;

}

// Function to get weight of an operator. An operator with higher weight will have higher precedence.

int GetOperatorWeight(char op)

{

int weight = -1;

switch(op)

{

case '+':

case '-':

weight = 1;

case '*':

case '/':

weight = 2;

case '$':

weight = 3;

}

return weight;

}

// Function to perform an operation and return output.

int HasHigherPrecedence(char op1, char op2)

{

int op1Weight = GetOperatorWeight(op1);

int op2Weight = GetOperatorWeight(op2);

// If operators have equal precedence, return true if they are left associative.

// return false, if right associative.

// if operator is left-associative, left one should be given priority.

if(op1Weight == op2Weight)

{

if(IsRightAssociative(op1)) return false;

else return true;

}

return op1Weight > op2Weight ? true: false;

}