How can you convert an infix notation to postfix notation?

Answer 1

#include <iostream>

#include <stack>

using namespace std;

int prec (char ch){

// Gives precedence to different operators

switch (ch) {

case '^':

return 5;

case '/':

return 4;

case '*':

return 4;

case '+':

return 2;

case '-':

return 1;

default :

return 0;

}

}

bool isOperand(char ch){

// Finds out is a character is an operand or not

if ((ch>='0' && ch<='9') (ch>='a' && ch<='z'))

return true;

else

return false;

}

string postFix (string infix){

string pfix = "";

stack<char> opstack;

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

// Scan character by character

if (isOperand(infix[i]))

{

pfix += infix[i];

}

else if (infix[i] ')')

{

// Retrace to last ( closure

while (opstack.top() != '(')

{

pfix += opstack.top();

opstack.pop();

}

// Remove the '(' found by while loop

opstack.pop();

}

Answer 2

If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.

To convert infix notation to postfix notation, parse the expression into tokens. As you encounter an operand, push it onto one stack. As you encounter an operator, consider its precedence level and, if it is higher than the current precedence level, push it and the precedence level onto a second stack. Note that the parenthesis is a special operator that overrides precedence level, raising it.

When you encounter an operator with a precedence level equal to or lower than the current precedence level, pop the operand(s) off of the first stack, output them, and the pop the operator off of the second stack (unless it was not pushed in the first place) and output it. As you pop things off of the stacks, consider the current precedence level against the current precedence level, and iterate the pop/output sequence as needed.

Answer 3

Possible. Here's an example:

(1+3)*(3+4)

1 3 + 3 4 + *

Answer 4

convert postfix notation to infix notation in c?

Answer 5

Both are expressions..

a+b is infix ...where operater comes between operands

ab+ is postfix..where operator comes at the last

Answer 6

Postfix is an open source software. It is used to setup mail servers.

Answer 7

(a*b+g/k)*l-(2*j)

Answer 8

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