Evaluation of postfix expression

this is just to explain how expression is converted from postfix to infix in data structure

1. DATA STRUCTURES USING ‘C’

2. Evaluation Of Postfix

3. Evaluation Of Postfix -BY AKHIL AHUJA

4. What is Postfix expression? Postfix expression is also known as Reverse Polish Notation(RPN). In RPN the operators follow their operands. IF THE OPERATOR SYMBOLS ARE PLACED AFTER ITS OPERANDS , THEN THE EXPRESSION IS IN POSTFIX NOTATION . Despite the name , Reverse Polish Notation is not exactly the reverse of polish notation , for the operands they are still written in conventional manner. For eg : “*63”in polish notation and “63*” in reverse polish notation. In this , expression is different but the answer is same (i.e 18)

5. Precedence of operators: 1. ‘$’ or ‘^’ Highest Precedence 2. ‘*’ and ‘/’ High Precedence 3. ’+’ and ‘-’ Low Precedence 4. ‘=‘ Lower Precedence 5. ‘(‘ and ‘)’ Lowest Precedence

6. Fundamental principles followed while evaluating the postfix expression : 1.Read the expression from left to right. 2.If there comes an operand , push it into the stack. 3.If there comes an operator , pop operand 1 and operand 2 and then : A . Push ‘(‘ B . Push ‘operand 2’ C . Push ‘operator’ D . Push ‘operand 1’ E . Push ‘)’

7.   (A). Postfix notation is easier to work with. In a postfix expression operators operands appear before the operators, there is no need of operator precedence and other rules. In postfix expression the topmost operands are popped off and operator is applied and the result is again pushed to the   stack and thus finally the stack contains a single value at the end of the process.   Advantages of Postfix Expression:  (B). In postfix expression, there are no parentheses and therefore the order of evaluation will be determined by the positions of the operators and related operands in the expression.

8. Algorithm for Postfix Expression : 1.Read the element. 2.If element is an operand then: Push the element into the stack. 3.If element is an operator then : Pop two operands from the stack. E valuate expression formed by two operand and the operator. Push the result of expression in the stack end . 4.If n o more elements then: Pop the result Else Goto step 1.

9. How to evaluate postfix (manually) Going from left to right, if you see an operator, apply it to the previous two operands (numbers) Example: Equivalent infix: A + B * C / D– (E– F) A B C * D / + E F - - (B*C) ((B*C )/ D) (A+(B*C)/D) (E – F) ((A+(B*C)/D)-(E-F))

10. Here’s an another example

11. 4 3 * 6 7 + 5 - + )

12. 4 3 * 6 7 + 5 - + ) STACK Push the opening bracket ‘(‘ into the stack . (

13. 4 3 * 6 7 + 5 - + ) STACK ( Operand ‘4’ push it into the stack 4

14. 4 3 * 6 7 + 5 - + ) STACK ( 4 Operand ‘3’ push it into the stack. 3

15. 4 3 * 6 7 + 5 - + ) STACK ( 4 3 * Now comes Operator ‘*’

16. 4 3 * 6 7 + 5 - + ) STACK ( 4 3 * Pop operand ‘3’ and ‘4’

17. 4 3 * 6 7 + 5 - + ) STACK ( 4 3 * Evaluate the expression 4*3 =12 Pus it into the stack

18. 4 3 * 6 7 + 5 - + ) STACK ( 4 3 * Evaluate the expression 4*3 =12 12

19. 4 3 * 6 7 + 5 - + ) STACK ( 4 3 * Push it into the stack 12

20. 4 3 * 6 7 + 5 - + ) STACK ( 12 Operand ‘6’ push it into the stack 6

21. 4 3 * 6 7 + 5 - + ) STACK ( 12 Operand ‘7’ push it into the stack 6 7

22. 4 3 * 6 7 + 5 - + ) STACK ( 12 Operand ‘+’ push it into the stack 6 7 + Operand ‘7’ and ‘6’ Are popped

23. 4 3 * 6 7 + 5 - + ) STACK ( 12 6 7 + Evaluate ‘6+7’ 13 =13

24. 4 3 * 6 7 + 5 - + ) STACK ( 12 13 Push it into the stack

25. 4 3 * 6 7 + 5 - + ) STACK ( 12 13 5 Operand ‘5’ push it into the stack

26. 4 3 * 6 7 + 5 - + ) STACK ( 12 13 5 Now operator ‘-’ occurs - Operands ‘5’ and ‘13’ are popped

27. 4 3 * 6 7 + 5 - + ) STACK ( 12 Evaluating “13-5” We get ‘8’ 8

28. 4 3 * 6 7 + 5 - + ) STACK ( 12 Now, push the resultant value i.e ‘8’ in the stack 8

29. 4 3 * 6 7 + 5 - + ) STACK ( 12 8 Operator ‘+’ +

30. 4 3 * 6 7 + 5 - + ) STACK ( 12 8 + By Evaluating , we get 20 20

31. 4 3 * 6 7 + 5 - + ) STACK ( So push ‘20’ in stack 20

32. 4 3 * 6 7 + 5 - + ) STACK ( 20 Now atlast ‘)’ occurs ) Now , pop the element till the opening bracket

33. 4 3 * 6 7 + 5 - + ) STACK ( 20 )

34. 4 3 * 6 7 + 5 - + ) STACK 20 Now push the element () AND THE FINAL ANSWRER IS ’20’

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