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# Create Postfix from infix experssion.

, 19 Aug 2006
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this article create postfix experssion from an infix experssion.
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## Introduction

Ali reza zareian.

This is a simple program that creates postfix expression from an infix expression. What is an infix expression?

(a+b)/d is a simple one. The infix expression is regular for mathematic. it means when u want to calculate this u must

Add a with b at the first time then divided with d.

If  we change this expression to postfix. We have this one ab+d/.

What is the useful of postfix?

At first we don't need parantes.Second we can write a program to calculate this expression easier than infix expression.

What is this program?

This programe use a simple algoritm to change infix to postfix.Just sees the example

Before I should say I use stack in this program.

Infix Program =(a+b)/d

Input     Stack        output

(             (

a            (                     a

+           (+                   a

b           (+                   ab

)                                  ab+

/            /                     ab+

d                                 ab+d

No         empty           ab+d/

Main method of this program!?

1)      private int isOperand(char  chrTemp);

2)      private int isOperator(char chrTemp);

3)      public string createPrifex();

1) isOperand check character if they are in this set.{'*','/','+','-','^','('}

If true we add this character to stack.

2)isOperator check character if they are in this set.{'*','/','+','-','^'}

This  method check the  stack  and  pop all the operator except '('.

3)createPrifix method create postfix from infix use isOperand and isOperator.

PesoCode

For (the first infix character ) to  last character

{

If  (is operand)   add to stack

Elae

If  (it is ')' )   pop  stack and send it to output

Else send other character to output

}

}

## License

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## About the Author

Software Developer
Iran (Islamic Republic Of)
Where is the truth?

## Comments and Discussions

 First Prev Next
 M_k m _k 16-Oct-13 7:01
 My vote of 1 Mohamad K Ayash 3-Oct-09 11:30
 Incorrect Solution naeem_libra 8-Dec-08 19:50
 Re: Incorrect Solution ali_reza_zareian 21-Jan-09 7:24
 Better method PIEBALDconsult 11-Jun-08 17:26
 WATCH OUT! This solution is wrong!! Member 4516628 17-Mar-08 13:14
 I am sorry, but your solution is not complete. U havent solved the precedence logic of the operations!   following step is needed:   If the scanned character is an Operator and the stack is not empty, compare the precedence of the character with the element on top of the stack. If top item has higher precedence over the scanned character Pop the stack else push the scanned character to stack. Repeat this step as long as stack is not empty and topStack has precedence over the character.
 Re: WATCH OUT! This solution is wrong!! ali_reza_zareian 18-Mar-08 9:14
 Re: WATCH OUT! This solution is wrong!! PKokkula 27-Jul-09 2:16
 plz help me behruoz guozali asl 8-Dec-07 7:01
 Re: plz help me ali_reza_zareian 21-Jan-09 7:07
 precedence logic is missing SoftwareCat 17-May-07 5:11
 Good? azam's 25-Feb-07 19:03
 Don't meant to complain but... Herbert Sauro 21-Aug-06 19:41
 Last Visit: 31-Dec-99 18:00     Last Update: 21-Sep-14 4:47 Refresh 1

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