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PPT – Infix Expression Evaluation PowerPoint presentation | free to view - id: 20d3b1-NzkyY

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Infix Expression Evaluation

Postfix expressions are easy to evaluate

But this is not the case for infix expressions!

e.g. 9 (2 3) 8

Two approaches to evaluate an infix expression

Infix Expression Attributes

Left associative , -, /,

Right associative

276 (3 2 4) 5

2(76) ((3 (2 4)) 5)

Decreasing order of precedence

() gt gt / gt

Rank of Expression

Binary operators only.

Evaluates an infix expression based on rank.

- for any operand
- -1 for , -, , /, ,
- 0 for (, )

Cumulative rank sum of the ranks of individual

terms.

2 7 6 ( 3 2 4 ) 5

cumulative rank

1 0 1 0 1 0 0 1 0 1 0 1 1 0 1

Cumulative rank for the entire expression is

always 1.

(exactly one more operand than operator)

Invalid expression if c.r. ? 1 (e.g. 2 4 3)

Infix-to-Postfix Conversion

During the scan of an expression

Operator stack

Example 1

The stack temporally stores operators awaiting

their right-hand-side operand.

a b c

has higher priority than ? add to the

stack

Operator Stack

Postfix string

a

c

b

Example 2

Use the stack to handle operators with same or

lower precedence.

a b / c d

has the same priority as / ? pop and

write it to the postfix string before

adding / to the stack.

/ has higher priority than

Operator Stack

/

Postfix string

a

b

/

c

d

Example 3

Use precedence values to handle (right

associative).

input precedence 4 when is the input. stack

precedence 3 when resides on the stack.

a b c

2nd has precedence 4 but 1st has only 3

? 2nd goes to operator stack (so it

will be popped before 1st )

Operator Stack

Postfix string

a

c

b

Example 4

Two precedence values for left parenthese (

input precedence 5 which is higher than that of

any operator. (all operators on the stack must

remain because a new subexpression begins.) stack

precedence -1 which is lower than that of any

opeartor. (no operator in the subexpression may

remove ( until ) shows up.)

a ( b c )

( has precedence 5 ? it goes to the

operator stack.

( now has precedence -1 ? it stays on the

operator stack.

Operator Stack

(

Postfix string

a

c

b

Input and Stack Precedence

Input Precedence

Stack Precedence

Rank

Symbol

- 1 1

-1 / 2

2 -1

4 3

-1 ( 5

-1 0 )

0 0

0

Rules for Evaluation

An Example

3 (4 2 5) 6

Operator stack

2

3

3

3

3 4

postfix

3 4

3 4 2

3 4 2

3 4 2 5

3 4 2 5 -

contd

Pop (

3 (4 2 5) 6

2

1

1

3 4 2 5 -

3 4 2 5 -

3 4 2 5 - 6

3 4 2 5 - 6

Class for the Symbols

Encapsulate each symbol along with the associated

input and output

precedence.

class expressionSymbol public expressionSymb

ol() // default constructor expressionSymbol(

char ch) // initializes the object for operator

ch friend bool operatorgt (const

expressionSymbol left,

const expressionSymbol right) // return true

if stackPrecedence of left is // gt

inputPrecedence of right. determines

whether // operator left on the stack should

be output before // pushing operator right on

the stack char getOp() const // return

operator private char op //

operator int inputPrecedence // input

precedence of op int stackPrecedence // stack

precedence of op

Class for Conversion

The infix2Postfix class uses a stack to store

expressionSymbol objects that correspond to

operators.

const char lParen '(', rParen ')' class

infix2Postfix public infix2Postfix() //

default constructor. infix expression is NULL

string infix2Postfix(const string

infixExp) // initialize the infix

expression void setInfixExp(const string

infixExp) // change the infix

expression string postfix() // return a

string that contains the equivalent postfix //

expression. the function throws expressionError

if an // error occurs during conversion

infix2Postfix contd

private string infixExpression // the

infix expression to convert string

postfixExpression // built to contain the

postfix equivalent of infixExpression stackltexpr

essionSymbolgt operatorStack // stack of

expressionSymbol objects void

outputHigherOrEqual(const expressionSymbol

op) // the expressionSymbol object op holds

the current // symbol. pop the stack and

output as long as the symbol // on the top of

the stack has a precedence gt that of // the

current operator bool isOperator(char ch)

const // is ch one of '','-','','/','',''

Output Stack Symbols

Pop symbols on the stack with stack precedence ?

input precendence of the new symbol.

void infix2PostfixoutputHigherOrEqual(const

expressionSymbol op) expressionSymbol

op2 while(!operatorStack.empty() (op2

operatorStack.top()) gt op) operatorStack.pop

() postfixExpression op2.getOp() postfixE

xpression ' '

Function for Infix-to-Postfix Conversion

postfix() does the following

Declarations

expressionSymbol op // maintain rank for error

checking int rank 0, i char ch

Processing an Operand

// process until end of the expression for

(i0 i lt infixExpression.length() i) ch

infixExpressioni // process an

operand // an operand is a single

digit non-negative integer if

(isdigit(ch)) // just add operand to

output expression, followed by // a

blank postfixExpression ch postfixExpres

sion ' ' // rank of an operand is 1,

accumulated rank // must be 1 rank if

(rank gt 1) throw expressionError("infix2P

ostfix Operator expected")

Processing an Operator

// process an operator or '('

else if (isOperator(ch) ch

'(') // rank of an operator is -1. rank of

'(' is 0. // accumulated rank should be

0 if (ch ! '(') rank-- if (rank lt

0) throw expressionError("infix2Postfix

Operand expected") else else //

output the operators on the stack with

higher // or equal precedence. push the

current operator // on the stack op

expressionSymbol(ch) outputHigherOrEqual(op)

operatorStack.push(op)

Processing a Right Parenthesis

else if (ch rParen) // build an

expressionSymbol object holding ')', which //

has precedence lower than the stack

precedence // of any operator except '('. pop

the operator stack // and output operators

from the subexpression until // '(' surfaces

or the stack is empty. if the stack is //

empty, a '(' is missing otherwise, pop off

'('. op expressionSymbol(ch) outputHigher

OrEqual(op) if (operatorStack.empty()) thr

ow expressionError("infix2Postfix Missing

'('") else operatorStack.pop() // get

rid of '(' // make sure ch is

whitespace else if

(!isspace(ch)) throw expressionError("infix2Pos

tfix Invalid input")

Finish Processing

// outside the for loop if (rank ! 1) throw

expressionError("infix2Postfix Operand

expected") else // flush operator stack

and complete expression evaluation. // if find

left parenthesis, a right parenthesis is

missing while (!operatorStack.empty()) op

operatorStack.top() operatorStack.pop()

if (op.getOp() lParen) throw

expressionError("infix2Postfix Missing

')'") else postfixExpression

op.getOp() postfixExpression '

' return postfixExpression

Evaluating an Infix Expression

void main() // use iexp for infix to postfix

conversion infix2Postfix iexp // infix

expression input and postfix expression

output string infixExp, postfixExp // use pexp

to evaluate postfix expressions postfixEval

pexp // input and evaluate infix expressions

until the // user enters an empty string //

get the first expression cout ltlt "Enter an infix

expression " getline(cin, infixExp) while

(infixExp ! "") // an exception may occur.

enclose the conversion // to postfix and the

output of the expression // value in a try

block

try // convert to postfix iexp.setInfi

xExp(infixExp) postfixExp

iexp.postfix() // output the postfix

expression cout ltlt "The postfix form is " ltlt

postfixExp ltlt endl // use pexp to evaluate

the postfix expression pexp.setPostfixExp(postf

ixExp) cout ltlt "Value of the expression

" ltlt pexp.evaluate() ltlt endl ltlt

endl // catch an exception and output the

error catch (const expressionError

ee) cout ltlt ee.what() ltlt endl ltlt

endl // input another expression cout ltlt

"Enter an infix expression " getline(cin,

infixExp)

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