Expression Evaluation: Expression BinaryVariableValue Never need an instance of Expression.
Evaluation of postfix expression
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DATA STRUCTURES USING ‘C’
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this is just to explain how expression is converted from postfix to infix in data structure
Transcript of Evaluation of postfix expression
DATA STRUCTURES USING
‘C’
Evaluation
Of
Postfix
Evaluation
Of
Postfix
-BY AKHIL AHUJA
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)
Precedence of operators:
1. ‘$’ or ‘^’ Highest Precedence
2. ‘*’ and ‘/’
High Precedence
3. ’+’ and ‘-’
Low Precedence
4. ‘=‘ Lower Precedence
5. ‘(‘ and ‘)’
Lowest Precedence
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 ‘)’
(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.
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.Evaluate expression formed by two
operand and the operator.Push the result of expression in the stack
end.4.If no more elements then:Pop the resultElseGoto step 1.
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))
Here’s an another example
4 3 * 6 7 + 5 - + )
4 3 * 6 7 + 5 - + )
STACK
Push the opening bracket ‘(‘ into the stack .
(
4 3 * 6 7 + 5 - + )
STACK
( Operand ‘4’ push it into the stack
4
4 3 * 6 7 + 5 - + )
STACK
(4 Operand ‘3’ push it into
the stack.
3
4 3 * 6 7 + 5 - + )
STACK
(43
*
Now comes Operator ‘*’
4 3 * 6 7 + 5 - + )
STACK
(
43
*
Pop operand ‘3’ and ‘4’
4 3 * 6 7 + 5 - + )
STACK
(
4
3*
Evaluate the expression4*3 =12Pus it into the stack
4 3 * 6 7 + 5 - + )
STACK
(
4
3*
Evaluate the expression4*3 =12
4 3 * 6 7 + 5 - + )
STACK
(
4
3*
Push it into the stack
12
4 3 * 6 7 + 5 - + )
STACK
(
12
Operand ‘6’ push it into the stack
6
4 3 * 6 7 + 5 - + )
STACK
(
12Operand ‘7’ push it into the stack
6
7
4 3 * 6 7 + 5 - + )
STACK
(
12Operand ‘+’ push it into the stack
67
+
Operand ‘7’ and ‘6’ Are popped
4 3 * 6 7 + 5 - + )
STACK
(
12
6 7+
Evaluate ‘6+7’
=13
4 3 * 6 7 + 5 - + )
STACK
(
12
13
Push it into the stack
4 3 * 6 7 + 5 - + )
STACK
(
1213
5
Operand ‘5’ push it into the stack
4 3 * 6 7 + 5 - + )
STACK
(
12135
Now operator ‘-’ occurs
-
Operands ‘5’ and ‘13’ are popped
4 3 * 6 7 + 5 - + )
STACK
(
12
Evaluating “13-5”
We get ‘8’
4 3 * 6 7 + 5 - + )
STACK
(
12Now, push the resultant value i.e ‘8’ in the stack
8
4 3 * 6 7 + 5 - + )
STACK
(
128
Operator ‘+’
+
4 3 * 6 7 + 5 - + )
STACK
(
12 8+
By Evaluating , we get 20
4 3 * 6 7 + 5 - + )
STACK
(
So push ‘20’ in stack
20
4 3 * 6 7 + 5 - + )
STACK
(
20Now atlast ‘)’ occurs
)
Now , pop the element till the opening bracket
4 3 * 6 7 + 5 - + )
STACK(
20
)
4 3 * 6 7 + 5 - + )
STACK
20
Now push the element
()AND THE FINAL ANSWRER IS ’20’
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