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Views: 2039 | Replies: 2
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Join Date: Nov 2006
Posts: 39
Reputation: 
Rep Power: 2
Solved Threads: 1
Light Poster
hey guys im supposed to write a simple calculator in java. Not in applet. Stack class has already been given and i was supposed to tokenize the input, have a function convert to take it from infix to postfix notation and eval to evaluate it. i have written the code convert but when i try to write the eval code i always get errors any idea how to start the eval function. my code is posted below
import java.util.*;
import java.math.*;
import java.io.*;
public class Tokens{
private int position=0; //Postion to put in array. Increases as the String is tokenized.
private String [] Tokenized= new String[500]; //Tokenized version of array
private Stack operand = new Stack(); //Operand Stack for conversion
private static Stack output=new Stack(); //Output Stack for conversion. (Everything will be in here at end of conversion)
public Tokens()
/********************************
*Start of tokenize class. Once called it will use the Parse function
*to tokenize the given String (ie. user input)*/
{
System.out.println("2+4-(5*3)^2^3+3");
Parse("2+4-(5*3)**2**3+3" + " ");
convert(Tokenized);
}
public void Parse(String s){
/****************************
*Parse Function. Takes Input as string and tokenizes it to be able to use as a Token.
****************************/
String temp=""; //Digits and operands are temporarily stored, seperatly in here.
for (int i=0; i<s.length();i++){
if(Character.isDigit(s.charAt(i))){
temp+=s.substring(i,i+1);
}
else
{
if(temp != "")
{
Tokenized[position]=temp;
position++;
}
temp="";
switch (s.charAt(i)){
case '+':
temp="+";
break;
case '-':
temp="-";
break;
case '(':
temp="(";
break;
case ')':
temp=")";
break;
case '/':
temp="/";
break;
case '*':
if (s.charAt(i+1) == '*')
{
temp="^";
i++;
}
else{
temp="*";
}
break;
case ' ':
break;
default:
System.out.println("Incorrect format.");
}
if(temp != "")
{
Tokenized[position]=temp;
position++;
}
temp = "";
}
}
}
public int evaluate(String x){
/*************************
Evaluates the given sign.
* +, - have same value:1
* *, / have same value:2
* ^ has value of: 3
* otherwise it is value: 0
*************************/
if(x.equals("+") )
{
return 1;
}
else if(x.equals("-"))
{
return 1;
}
else if(x.equals("*"))
{
return 2;
}
else if (x.equals("/"))
{
return 2;
}
else if(x.equals("^"))
{
return 3;
}
else
{
return 0;
}
}
public void Bracket()
{
String temp=((String)(operand.peek()));
while(!((String)(operand.peek())).equals("("))
{
output.push(operand.pop());
}
operand.pop();
}
public void convert(String[] Tokenized){
/****************************************************
*takes the token created in Parse function and
*converts it from Infix notation to postfix notion
*Uses evaluate function to see what to do with signs
* and puts it in 2 stacks operand and output
*untill the end. at the end everything is in
*stack output which will be returned
****************************************************/
for (int j=0; j<position; j++)
{
if (Character.isDigit(Tokenized[j].charAt(0)))
{
output.push(Tokenized[j]);
}
else
{
if (!Tokenized[j].equals("("))
{
if (operand.empty())
{
operand.push(Tokenized[j]);
}
else if (((String)(operand.peek())).equals("("))
{
operand.push(Tokenized[j]);
}
else if (Tokenized[j].equals(")"))
{
Bracket();
}
else if (Tokenized[j].equals("^"))
{
if (((String)(operand.peek())).equals("^"))
{
operand.push(Tokenized[j]);
}
else
{
int second=evaluate((String)(operand.peek()));
int first=evaluate(Tokenized[j]);
if (first > second)
{
operand.push(Tokenized[j]);
}
else if (first == second)
{
output.push(operand.pop());
operand.push(Tokenized[j]);
}
else if (first < second)
{
output.push(operand.pop());
operand.push(Tokenized[j]);
}
}
}
else
{
int second=evaluate((String)(operand.peek()));
int first=evaluate(Tokenized[j]);
if (first > second)
{
operand.push(Tokenized[j]);
}
else if (first == second)
{
output.push(operand.pop());
operand.push(Tokenized[j]);
}
else if (first < second)
{
output.push(operand.pop());
operand.push(Tokenized[j]);
}
}
}
else if (Tokenized[j].equals("("))
{
operand.push(Tokenized[j]);
}
}
}
while(!operand.empty())
{
output.push(operand.pop());
}
}
public static void main(String[] args){
Tokens app = new Tokens();
}
} 
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Join Date: Nov 2006
Posts: 39
Reputation:
Rep Power: 2
Solved Threads: 1
Light Poster
k guys i got somewhat how to do the eval but does someone mind looking at the parse function for me. it is supposed to tokenize a string that is written for a calc such as 3+4/(5+1) but it will not take the last entry unless i add a space " " to it. any1 have a fix for it? my program runs fine like it is but it is because i manipulated it to. any ideas?
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Join Date: Aug 2005
Posts: 4,663
Reputation:
Rep Power: 16
Solved Threads: 297
>2+4-(5*3)**2**3+3"
That is not a valid expression. Personally I'd chop that double
After conversion from infix to postfix, would be:-
Thus working from left to right...
2 4 + 5 3 * 2 2 ^ ^ - 3 +
2 + 4 = 6 {pop}
6 5 3 * 2 2 ^ ^ - 3 +
5 * 3 = 15 {pop}
6 15 2 2 ^ ^ - 3 +
2 ^ 2 = 4 {pop}
6 15 4 ^ - 3 +
15 ^ 4 = 50625 {pop}
6 50625 - 3 +
6 - 50625 = -50619 {pop}
-50619 3 +
-50619 + 3 = -50616
There's noffin left to pop so that's your answer.
Look a my sample snippet->http://www.daniweb.com/code/snippet599.html
That is not a valid expression. Personally I'd chop that double
** out.2+4-(5*3)^2^3+3
After conversion from infix to postfix, would be:-
2 4 + 5 3 * 2 2 ^ ^ - 3 +
Thus working from left to right...
2 4 + 5 3 * 2 2 ^ ^ - 3 +
2 + 4 = 6 {pop}
6 5 3 * 2 2 ^ ^ - 3 +
5 * 3 = 15 {pop}
6 15 2 2 ^ ^ - 3 +
2 ^ 2 = 4 {pop}
6 15 4 ^ - 3 +
15 ^ 4 = 50625 {pop}
6 50625 - 3 +
6 - 50625 = -50619 {pop}
-50619 3 +
-50619 + 3 = -50616
There's noffin left to pop so that's your answer.
Look a my sample snippet->http://www.daniweb.com/code/snippet599.html
Last edited by iamthwee : Mar 24th, 2007 at 4:41 am.
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