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Infix to postfix converstion using stack

Poised Pig answered on May 2, 2021 Popularity 10/10 Helpfulness 2/10

answer Infix to postfix converstion using stack

Infix to postfix converstion using stack

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/*
  Infix to postfix conversion in C++ 
  Input Postfix expression must be in a desired format. 
  Operands and operator, both must be single character.
  Only '+'  ,  '-'  , '*', '/' and '$' (for exponentiation)  operators are expected. 
*/
#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;
}

Popularity 10/10 Helpfulness 2/10 Language whatever

Source: gist.githubusercontent.com

Tags: postfix stack using whatever

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Contributed on May 02 2021

Poised Pig

0 Answers Avg Quality 2/10

Infix to postfix converstion using stack

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