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Posted 2 Nov 2003

# Expression evaluator : using RPN

, 5 Nov 2003
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An article showing how to evaluate mathematical expressions using reverse polish notation (RPN)

## Introduction

This article will demonstrate how to evaluate complex mathematical expressions by converting them from infix notation to postfix notation and evaluating the expression. In the process we will be using STL's stack and string classes. When finished, the program should be able to evaluate expressions such as:

`expr = "(12232+(43*43-(250/(3*8))*44)/12-311) * (5==5)"`

## Background

Most programming languages require that you enter expressions in infix notation that is: operators and operands are intermixed, example: "`5*6-3*2*3`".

The postfix notation, introduced in 1950 by the Polish logician Jan Lukasiewicz, is a method of representing an expression without using parenthesis and still conserving the precedence rules of the original expression. For example, the previous expression could have been written like: "`5 6 * 3 2 * 3 * -`"

## Explaining the code

Here's is how to convert from an infix notation to postfix notation:

1. Initialize an empty stack (string stack), prepare input infix expression and clear RPN string
2. Repeat until we reach end of infix expression
1. Get token (operand or operator); skip white spaces
2. If token is:
1. Left parenthesis: Push it into stack
2. Right parenthesis: Keep popping from the stack and appending to RPN string until we reach the left parenthesis.
If stack becomes empty and we didn't reach the left parenthesis then break out with error "Unbalanced parenthesis"
3. Operator: If stack is empty or operator has a higher precedence than the top of the stack then push operator into stack. Else if operator has lower precedence then we keep popping and appending to RPN string, this is repeated until operator in stack has lower precedence than the current operator.
4. An operand: we simply append it to RPN string.
3. When the infix expression is finished, we start popping off the stack and appending to RPN string till stack becomes empty.

Now evaluating a postfix (RPN) expression is even easier:

1. Initialize stack (integer stack) for storing results, prepare input postfix (or RPN) expression.
2. Start scanning from left to right till we reach end of RPN expression
3. Get token, if token is:
1. An operator:
1. Get top of stack and store into variable op2; Pop the stack
2. Get top of stack and store into variable op1; Pop the stack
3. Do the operation expression in operator on both op1 and op2
4. Push the result into the stack
2. An operand: stack its numerical representation into our numerical stack.
4. At the end of the RPN expression, the stack should only have one value and that should be the result and can be retrieved from the top of the stack.
To use the code:
```#include <iostream.h>
#include <string>
#include "ExpressionEvaluator.h"

using std::string;

int main()
{
long result;
double resultdbl;
int err;

string s;

s = "1+2*(1-2-3-4)";
err = ExpressionEvaluator::calculateLong(s, result);
if (err != ExpressionEvaluator::eval_ok)
cout << "Error while evaluating!" << endl;
else
cout << "Evaluation of (int):" << s.c_str() << " yielded: "
<< result << endl;

s = "1.1/5.5+99-(4.1*(2+1)-5)";
err = ExpressionEvaluator::calculateDouble(s, resultdbl);
if (err != ExpressionEvaluator::eval_ok)
cout << "Error while evaluating!" << endl;
else
cout << "Evaluation of (double):" << s.c_str() << " yielded: "
<< resultdbl << endl;

return 0;
}```

## Extending the code

This code can be extended to allow you perform other operations, however they must be binary operation (takes two operands). To extended the code simply add a new operator into the "operators" array along with its precedence value. If you introduce a new symbol make sure you add the symbol into the "operators[0]" string too. Precedence is important for generating a proper postfix expression. After adding a new operator, define its behaviour in the "`evaluateRPN`" function as:

```if (token == "PUT YOUR OPERATOR SYMBOL HERE")
r = doMyOperation(op1, op2);
```

Hope you find this code and article useful.

## History

• Sunday, November 2, 2003
• Initial version
• Monday, November 3, 2003
• Fixed precedence rule of multiplication
• Tuesday, November 4, 2003
• Fixed a bug in` isOperator() `
• Added support for negative and positive numbers as: -1 or +1
(initially they were supported as: 0-1 or 0+1)
• Added exception handling and foolproof against malformed expression
• Added >=, <=, != operators

A list of licenses authors might use can be found here

## Share

 Web Developer United States
Elias (aka lallousx86, @0xeb) has always been interested in the making of things and their inner workings.

His computer interests include system programming, reverse engineering, writing libraries, tutorials and articles.

In his free time, and apart from researching, his favorite reading topics include: dreams, metaphysics, philosophy, psychology and any other human/mystical science.

Former employee of Hex-Rays (the creators of IDA Pro), was responsible about many debugger plugins, IDAPython project ownership and what not.

Elias currently works at Microsoft as a software security engineer.

More articles and blog posts can be found here:

- http://lallousx86.wordpress.com/
- http://0xeb.wordpress.com/
- http://www.hexblog.com/?author=3

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 NOT operator? twinbee4-Jun-08 8:16 twinbee 4-Jun-08 8:16
 How about to offer a unicode version? Peter, Chan12-Jul-07 16:38 Peter, Chan 12-Jul-07 16:38
 Re: How about to offer a unicode version? lallous15-Jul-07 23:26 lallous 15-Jul-07 23:26
 negative numbers after operators banjaxx12-Nov-06 4:29 banjaxx 12-Nov-06 4:29
 Re: negative numbers after operators lallous14-Nov-06 22:42 lallous 14-Nov-06 22:42
 Re: negative numbers after operators banjaxx16-Nov-06 20:18 banjaxx 16-Nov-06 20:18
 Re: negative numbers after operators agahowa11-May-10 6:22 agahowa 11-May-10 6:22
 Thanks to the author jewelgal10-Apr-06 15:31 jewelgal 10-Apr-06 15:31
 Re: Thanks to the author lallous10-Apr-06 20:50 lallous 10-Apr-06 20:50
 Re: Thanks to the author jewelgal11-Apr-06 21:49 jewelgal 11-Apr-06 21:49
 Re: Thanks to the author lallous11-Apr-06 22:03 lallous 11-Apr-06 22:03
 Thanks! (and get rid of std::string) cccMangus8-Nov-05 10:16 cccMangus 8-Nov-05 10:16
 Re: Thanks! (and get rid of std::string) lallous10-Nov-05 23:05 lallous 10-Nov-05 23:05
 Re: Thanks! (and get rid of std::string) Rusty FunkNut13-Aug-06 3:31 Rusty FunkNut 13-Aug-06 3:31
 != operator arisnova19-Jan-05 7:49 arisnova 19-Jan-05 7:49
 Re: != operator lallous19-Jan-05 22:26 lallous 19-Jan-05 22:26
 negation and exponentiation Anonymous6-Sep-04 18:52 Anonymous 6-Sep-04 18:52
 I've been trying to figure out a way to handle cases such as these: -2^2=-4 and 2^-2=0.25 In the first case the exponentiation operator has precedence but in the second case the negation operator has precedence. This must be handled when converting from infix to postfix. Is there a simple solution to this problem. There are, of course, many ugly ways to solve this--just hoping for one of those solutions that makes me say "Now why didn't I think of that?"
 Re: negation and exponentiation lallous7-Sep-04 21:04 lallous 7-Sep-04 21:04
 Re: negation and exponentiation David Orel11-Sep-04 18:16 David Orel 11-Sep-04 18:16
 Cool! However, it doesn't take floating point input! cccMangus14-Aug-04 21:03 cccMangus 14-Aug-04 21:03
 Re: Cool! However, it doesn't take floating point input! lallous17-Aug-04 21:05 lallous 17-Aug-04 21:05
 Re: Cool! However, it doesn't take floating point input! cccMangus18-Aug-04 2:57 cccMangus 18-Aug-04 2:57
 How about sin Isidor18-Jan-04 7:28 Isidor 18-Jan-04 7:28
 How about the OPs that only take one operand? CatGor28-Dec-03 23:30 CatGor 28-Dec-03 23:30
 Re: How about the OPs that only take one operand? Stephane Rodriguez.29-Dec-03 0:59 Stephane Rodriguez. 29-Dec-03 0:59
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