Stacks & Queues Infix Calculator

1. Stacks & QueuesInfix Calculator CSC 172 SPRING 2002 LECTURE 5

2. Workshop sign-up Still time : Dave Feil-Seifer df001i@mail.rochester.edu Ross Carmara rc001i@mail.rochester.edu

3. Infix to postfix 1 + 2 * 3 == 7 (because multiplication has higher precidence) 10 – 4 – 3 == 3 (because subtraction proceeds left to right)

4. Infix to postfix 4 ^ 3 ^ 2 == 262144 != 4096 Generally, Rather than:

5. Precidence A few simple rules: () > ^ > * / > + - Subtraction associates left-to-right Exponentiation associates right to left

6. Infix Evaluation 1 – 2 – 4 ^ 5 * 3 * 6 / 7 ^ 2 ^ 2 == -8 (1 – 2) – ( ( ( ( 4 ^ 5) * 3) * 6) / (7 ^ ( 2 ^ 2 ) ) ) Could you write a program to evaluate stuff like this?

7. Postfix • If we expressed (1 – 2) – ( ( ( ( 4 ^ 5) * 3) * 6) / (7 ^ ( 2 ^ 2 ) ) ) As 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - Then, we could use the postfix stack evaluator

8. Postfix evaluation using a stack • Make an empty stack • Read tokens until EOF • If operand push onto stack • If operator • Pop two stack values • Perform binary operation • Push result • At EOF, pop final result

9. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -

10. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 1

11. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 2 1

12. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - -1

13. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 4 -1

14. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 5 4 -1

15. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 1024 -1

16. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 3 1024 -1

17. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 3072 -1

18. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 6 3072 -1

19. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 18432 -1

20. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 7 18432 -1

21. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 2 7 18432 -1

22. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 2 2 7 18432 -1

23. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 4 7 18432 -1

24. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 2041 18432 -1

25. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - 7 -1

26. 1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / - -8

27. But how to go from infix to postfix? • Could you write a program to do it? • What data structures would you use • Stack • Queue • How about a simple case just using “+” • 1+ 2 + 7 + 4 • 1 2 7 4 + + + • Operands send on to output? • Operator push on stack? • Pop ‘em all at the end?

28. More complex 2 ^ 5 – 1 == 2 5 ^ 1 – Modify the simple rule? If you are an operator, pop first, then push yourself? 1 + 2 + 7 + 4 1 2 + 7 + 4 + ok

29. Even more complex 3 * 2 ^ 5 - 1 3 2 5 ^ * 1 – If you are an operator: Pop if the top of the stack is higher precedence than

30. Infix to postfix Stack Algorithm Operands : Immediately output Close parenthesis: Pop stack until open parenthesis Operators: • Pop all stack symbols until a symbol of lower precedence (or a right-associative symbol of equal precedence) appears. • Push operator EOF: pop all remaining stack symbols

31. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7

32. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 1

33. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 - 1

34. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 - 1 2

35. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 ^ - 1 2

36. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 ^ - 1 2 3

37. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 ^ ^ - 1 2 3

38. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 ^ ^ - 1 2 3 3

39. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 - 1 2 3 3 ^ ^ -

40. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 ( - 1 2 3 3 ^ ^ -

41. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 ( - 1 2 3 3 ^ ^ - 4

42. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 + ( - 1 2 3 3 ^ ^ - 4

43. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 + ( - 1 2 3 3 ^ ^ - 4 5

44. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 * + ( - 1 2 3 3 ^ ^ - 4 5

45. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 * + ( - 1 2 3 3 ^ ^ - 4 5 6

46. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 - 1 2 3 3 ^ ^ - 4 5 6 * +

47. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 * - 1 2 3 3 ^ ^ - 4 5 6 * +

48. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 * - 1 2 3 3 ^ ^ - 4 5 6 * + 7

49. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 1 2 3 3 ^ ^ - 4 5 6 * + 7 * -

50. 1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7 ((1 – (2 ^ (3 ^ 3))) – (( 4 + (5 * 6)) * 7)) To evaluation stack Input 1 2 3 3 ^ ^ - 4 5 6 * + 7 * -