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Simple Guide to Mathematical Expression Parsing

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18 Jun 2010 1  
In this article, I introduce a very simple way to parse an expression.

This article is dedicated to my love Aida.

Introduction

This article describes a practical mathematical parser - an interactive program that can be used to perform basic arithmetic operations. In fact, in order to keep understanding of this article simple, I prefer to avoid any unnecessary items so you cannot rely on this program as a real calculator and it just works with integers and it can parse an expression with + , - , * , / , ( , ).

When I first started thinking about writing code to calculate a mathematical expression, I soon realized that it wasn't that easy [at least for me]. I had to delve into such concepts as parsing, tokens, lexical analysis, node trees and tables, etc.

The rules of the algorithm were the same ones we learned in algebra. For example: multiply (or divide) before adding or subtracting; start with inner parentheses and work outwards; if a pair of binary operators have equal precedence, perform the left operator of the pair first.

Here is its screen shot:

csharpcalc.jpg

Using the Code

The idea of this calculator or better to say math parser is think in tree and node way. I think the following picture can explain the whole idea:

a+b*c*(d-e)

img004.gif

You can see the algorithm of multiplying (or dividing) before adding or subtracting; starting with inner parentheses and working outwards; if a pair of binary operators have equal precedence, performing the left operator of the pair first.

At first, I made ParseExpr() function. This function + or - two operands with each other but before that checks if there is any * or / after each operand or not so it calls ParseFactor() function which is made to * or / two operands. Of course it checks if there is any expression in a pair of parenthesis or not, so it calls ParseTerm() function to check those expressions between braces, then at this point it reaches the value so we should parse it so then it calls ParseNumber().

For example above at first it calls ParseExpr(Expression) -> ParseFactor(Expression) -> ParseTerm(Expression) -> ParseNumber(Expression) and it returns a and then it cuts the Expression to +b*c*(d-e) so the returned value from ParseNumber goes to op and then it checks the first character to see if it is plus and there it is, so it cuts the Expression to b*c*(d-e) and then it repeats the last actions and makes up the above tree.

Blocks of code:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace csharpcalc
{
    class Expression
    {
        public int ParseExpr(ref string expr)
        {
            int op , op1;
            op = ParseFactor(ref expr);
            if (expr.Length != 0 )
            {
                if (expr[0] == '+')
                {
                    expr = expr.Substring(1, expr.Length - 1);
                    op1 = ParseExpr(ref expr);
                    op += op1;
                }
                else if (expr[0] == '-')
                {
                    expr = expr.Substring(1, expr.Length - 1);
                    op1 = ParseExpr(ref expr);
                    op -= op1;
                }
            }
            return op;
        }
        public int ParseFactor(ref string expr)
        {
            int op, op1;
            op = ParseTerm(ref expr);
            if (expr.Length != 0)
            {
                if (expr[0] == '*')
                {
                    expr = expr.Substring(1, expr.Length - 1);
                    op1 = ParseFactor(ref expr);
                    op *= op1;
                }
                else if (expr[0] == '/')
                {
                    expr = expr.Substring(1, expr.Length - 1);
                    op1 = ParseFactor(ref expr);
                    op /= op1;
                }
            }
            return op;
         }
        public int ParseTerm(ref string expr)
        {
            int returnValue = 0;
            if (expr.Length != 0)
            {
                if (char.IsDigit(expr[0]))
                { 
                    returnValue = ParseNumber(ref expr);
                    return returnValue;
                }
                else if (expr[0] == '(')
                {
                    expr = expr.Substring(1, expr.Length - 1);
                    returnValue = ParseExpr(ref expr);
                    return returnValue;
                }
                else if (expr[0] == ')')
                    expr = expr.Substring(1, expr.Length - 1);
            }
            return returnValue;
        }
        public int ParseNumber(ref string expr)
        {
            string numberTemp = "";
            for (int i = 0; i < expr.Length && char.IsDigit(expr[i]); i++)
            {
                if (char.IsDigit(expr[0]))
                {
                    numberTemp += expr[0];
                    expr = expr.Substring(1, expr.Length - 1);
                }
            }
            return int.Parse(numberTemp);
        }
    }
}

If You Want To ...

If you want to complete this project in order to accept double numbers too, you can add if-block in ParseNumber() method that even if char.IsDigit() was false, check if it is "." and if it is so, add it to the numberTemp and then cast it.

If you want to add some other mathematical functions to this project such as sin(), cos(), etc., you can check the first char of Expression in the ParseTerm function to see if it is a letter or not (char.IsLetter) and then call a function like ParseWord() and thereafter splitting the word, you can use a switch case block.

Points of Interest

I am a Control Engineer, but I have always loved computer programming and this simple project brings life to my past dreams.

History

Since I was a 16 year old student, I had a dream of writing a math parser but I have always had some other work to do. Finally I decided to write that. After searching the literature, I came across an outline of an algorithm for parsing arithmetic expressions in Principles of Systems Programming by Robert M. Graham (Wiley, 1975).

License

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