Stacks & Queues Infix Calculator CSC 172 SPRING 2002 LECTURE 5.
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Stacks & Queues Infix Calculator CSC 172 SPRING 2002 LECTURE 5
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Transcript of Stacks & Queues Infix Calculator CSC 172 SPRING 2002 LECTURE 5.
Stacks & QueuesInfix Calculator
CSC 172
SPRING 2002
LECTURE 5
Infix to postfix
1 + 2 * 3
== 7 (because multiplication has higher precidence)
10 – 4 – 3
== 3 (because subtraction proceeds left to right)
Infix to postfix
4 ^ 3 ^ 2== 262144!= 4096
234
234
Generally,
Rather than:
Precidence
A few simple rules:
() > ^ > * / > + -
Subtraction associates left-to-right
Exponentiation associates right to left
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?
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
Postfix evaluation using a stack
1. Make an empty stack
2. Read tokens until EOFa. If operand push onto stack
b. If operator i. Pop two stack values
ii. Perform binary operation
iii. Push result
3. At EOF, pop final result
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
21
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
4-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
5 4-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
1024-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
31024
-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
3072-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
63072
-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
18432-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
718432
-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
27
18432-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
227
18432-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
47
18432-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
204118432
-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
7-1
1 2 – 4 5 ^ 3 * 6 * 7 2 2 ^ ^ / -
-8
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?
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
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
Infix to postfix Stack AlgorithmOperands : Immediately output
Close parenthesis: Pop stack until open parenthesis
Operators: 1. Pop all stack symbols until a symbol of lower
precedence (or a right-associative symbol of equal precedence) appears.
2. Push operator
EOF: pop all remaining stack symbols
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
1
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
-
1
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
-
1 2
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
^-
1 2
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
^-
1 2 3
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
^^-
1 2 3
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
^^-
1 2 3 3
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
-
1 2 3 3 ^ ^ -
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
(-
1 2 3 3 ^ ^ -
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
(-
1 2 3 3 ^ ^ - 4
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
+(-
1 2 3 3 ^ ^ - 4
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
+(-
1 2 3 3 ^ ^ - 4 5
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
*+(-
1 2 3 3 ^ ^ - 4 5
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
*+(-
1 2 3 3 ^ ^ - 4 5 6
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
-
1 2 3 3 ^ ^ - 4 5 6 * +
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
*-
1 2 3 3 ^ ^ - 4 5 6 * +
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
*-
1 2 3 3 ^ ^ - 4 5 6 * + 7
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
1 2 3 3 ^ ^ - 4 5 6 * + 7 * -
1 – 2 ^ 3 ^ 3 – ( 4 + 5 * 6) * 7
1 2 3 3 ^ ^ - 4 5 6 * + 7 * -
((1 – (2 ^ (3 ^ 3))) – (( 4 + (5 * 6)) * 7))
InputTo evaluationstack
