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infix to postfix in c++

i200825 Hammad Hassan Khan answered on January 6, 2022 Popularity 10/10 Helpfulness 2/10

answer infix to postfix in c++

infix to postfix in c++

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#include<iostream>
#include<stack>
using namespace std;

bool isOperator(char c)
{
	if(c=='+'||c=='-'||c=='*'||c=='/'||c=='^')
	{
		return true;
	}
	else
	{
		return false;
	}
}

int precedence(char c) 
{ 
    if(c == '^') 
    return 3; 
    else if(c == '*' || c == '/') 
    return 2; 
    else if(c == '+' || c == '-') 
    return 1; 
    else
    return -1; 
} 

string InfixToPostfix(stack<char> s, string infix)
{
	string postfix;
	for(int i=0;i<infix.length();i++)
	{
		if((infix[i] >= 'a' && infix[i] <= 'z')
		||(infix[i] >= 'A' && infix[i] <= 'Z'))
		{
			postfix+=infix[i];
		}
		else if(infix[i] == '(')
		{
			s.push(infix[i]);
		}
		else if(infix[i] == ')')
		{
			while((s.top()!='(') && (!s.empty()))
			{
				char temp=s.top();
				postfix+=temp;
				s.pop();
			}
			if(s.top()=='(')
			{
				s.pop();
			}
		}
		else if(isOperator(infix[i]))
		{
			if(s.empty())
			{
				s.push(infix[i]);
			}
			else
			{
				if(precedence(infix[i])>precedence(s.top()))
				{
					s.push(infix[i]);
				}	
				else if((precedence(infix[i])==precedence(s.top()))&&(infix[i]=='^'))
				{
					s.push(infix[i]);
				}
				else
				{
					while((!s.empty())&&( precedence(infix[i])<=precedence(s.top())))
					{
						postfix+=s.top();
						s.pop();
					}
					s.push(infix[i]);
				}
			}
		}
	}
	while(!s.empty())
	{
		postfix+=s.top();
		s.pop();
	}
	
	return postfix;
}

int main() 
{  

  	string infix_exp, postfix_exp;
  	cout<<"Enter a Infix Expression :"<<endl;
  	cin>>infix_exp;
  	stack <char> stack;
	cout<<"INFIX EXPRESSION: "<<infix_exp<<endl;
  	postfix_exp = InfixToPostfix(stack, infix_exp);
  	cout<<endl<<"POSTFIX EXPRESSION: "<<postfix_exp;
	  
	return 0;
}

Popularity 10/10 Helpfulness 2/10 Language cpp

Source: simplesnippets.tech

Tags: c++ postfix

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Contributed on Nov 15 2022

i200825 Hammad Hassan Khan

0 Answers Avg Quality 2/10

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cpp infix to postfix.

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//aight bru, lessgo. 
// use this exammple->(a-b/c)*(a/k-l) and implent code on paper to get a good idea.
#include<iostream>
#include<stack>
using namespace std;
int prec(char c)
{
	if(c=='^') return 3;
	else if(c=='*' || c=='/') return 2;
	else if(c=='+' || c=='-') return 1;
	else 
	return -1; // for open brackets. coz we have to pop until we find an opening bracket.
}
string infixtopostfix(string s)
{
	stack<char>st;
	string res="";
	for(int i=0;i<s.length();i++)
	{
		if((s[i]>='a' && s[i]<='z')||(s[i]>='A' && s[i]<='Z')) res=res+s[i];
		else if(s[i]=='(') st.push(s[i]);
		else if(s[i]==')') 
		{
			while(!st.empty() && st.top()!='(')
			{
				res=res+st.top();
				st.pop();
			}
			if(!st.empty()) st.pop();
		}
		else 
		{
			while(!st.empty() && prec(st.top())>prec(s[i]) )
			{
				res=res+st.top();
				st.pop();
			}
			st.push(s[i]);
		}
	}
	while(!st.empty())
	{
		res=res+st.top();
		st.pop();
	} 
	return res;
}
int main()
{
    string s;
    cin>>s;
	cout<<infixtopostfix(s);
	return 0;
}

Popularity 10/10 Helpfulness 5/10 Language cpp

Source: Grepper

Tags: c++

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Contributed on Jul 29 2022

ROCKETMAN

0 Answers Avg Quality 2/10

postfix to infix c program

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#include<stdio.h>
int main()
{
	char str[100],oprator[100],operant[100];
    int i ,opr=-1,op=-1;
    printf("enter the string");
    scanf("%s",str);
    for (i=0;str[i]!=NULL;i++)
    {
    	if(str[i]=='+'||str[i]=='-'||str[i]=='*'||str[i]=='^'||str[i]=='/')
    	{
    	(opr)++;
    		oprator[opr]=str[i];
		}
		else
		{
			(op)++;
    		operant[op]=str[i];
		}
    	
	}
     printf("operator are:\n");
     for (i=0;i<=opr;i++)
     printf("%c",oprator[i]);
     printf("\noperandas are:\n");
    for (i=0;i<=op;i++)
    printf("%c",operant[i]);
    return 0;
}

Popularity 9/10 Helpfulness 5/10 Language whatever

Source: Grepper

Tags: c postfix po

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Contributed on Jan 06 2022

Tired Toad

0 Answers Avg Quality 2/10

infix to postfix

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// C++ code to convert infix expression to postfix
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to return precedence of operators
int prec(char c)
{
    if (c == '^')
        return 3;
    else if (c == '/' || c == '*')
        return 2;
    else if (c == '+' || c == '-')
        return 1;
    else
        return -1;
}
  
// The main function to convert infix expression
// to postfix expression
void infixToPostfix(string s)
{
  
    stack<char> st;
    string result;
  
    for (int i = 0; i < s.length(); i++) {
        char c = s[i];
  
        // If the scanned character is
        // an operand, add it to output string.
        if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')
            || (c >= '0' && c <= '9'))
            result += c;
  
        // If the scanned character is an
        // ‘(‘, push it to the stack.
        else if (c == '(')
            st.push('(');
  
        // If the scanned character is an ‘)’,
        // pop and add to output string from the stack
        // until an ‘(‘ is encountered.
        else if (c == ')') {
            while (st.top() != '(') {
                result += st.top();
                st.pop();
            }
            st.pop();
        }
  
        // If an operator is scanned
        else {
            while (!st.empty()
                   && prec(s[i]) <= prec(st.top())) {
                result += st.top();
                st.pop();
            }
            st.push(c);
        }
    }
  
    // Pop all the remaining elements from the stack
    while (!st.empty()) {
        result += st.top();
        st.pop();
    }
  
    cout << result << endl;
}
  
// Driver code
int main()
{
    string exp = "a+b*(c^d-e)^(f+g*h)-i";
  
    // Function call
    infixToPostfix(exp);
    
    return 0;
}

Popularity 10/10 Helpfulness 2/10 Language whatever

Source: www.geeksforgeeks.org

Tags: postfix

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Contributed on Jul 22 2023

Azad Mahmud

0 Answers Avg Quality 2/10

infix to postfix in c++

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