Evaluate the infix expression with stack in one pass
Algorithm
Input:An arithmetic expression P in postfix notation;
1. scan P left to right; if character is operand push in stack.
2. else if character is operator
i. while the top of the stack is not of smaller precedence than this character.
ii. pop from stack.
iii. store let first operand operator second operand on the stack and go to 2.
3. pop operators till operator stack is not empty.
4. pop top 2 operands and push first operand operator second operand on the stack.
5. exit.
