Solving simple string expressions (1+2*3) in Java [Almost Done] -

i'm trying solve simple string expressions e.g. 1+2*3/4, without brackets. i'm done simple integer part, above expression work perfectly, i'm stuck decimal values, example 1.1/2.2*4.4 want push whole decimal number stack (double), i'm working on quite time not quite getting it, appreciated. current code is:

import java.util.stack; import java.text.decimalformat;  public class evaluatestring {     public static double evaluate(string expression)     {         char[] tokens = expression.tochararray();         decimalformat df = new decimalformat("#.##");           // stack numbers: 'values'         stack<double> values = new stack<double>();          // stack operators: 'ops'         stack<character> ops = new stack<character>();          (int = 0; < tokens.length; i++)         {              // current token whitespace, skip             if (tokens[i] == ' ')                 continue;              // current token number, push stack numbers             if (tokens[i] >= '0' && tokens[i] <= '9')             {                 stringbuffer sbuf = new stringbuffer();                 // there may more 1 digits in number                 while (i < tokens.length && tokens[i] >= '0' && tokens[i] <= '9')                     sbuf.append(tokens[i++]);                 values.push(double.parsedouble(sbuf.tostring()));             }              // current token operator.             else if (tokens[i] == '+' || tokens[i] == '-' ||                      tokens[i] == '*' || tokens[i] == '/')             {                 // while top of 'ops' has same or greater precedence current                 // token, operator. apply operator on top of 'ops'                 // top 2 elements in values stack                 while (!ops.empty() && hasprecedence(tokens[i], ops.peek()))                   values.push(applyop(ops.pop(), values.pop(), values.pop()));                  // push current token 'ops'.                 ops.push(tokens[i]);             }         }          // entire expression has been parsed @ point, apply remaining         // ops remaining values         while (!ops.empty())             values.push(applyop(ops.pop(), values.pop(), values.pop()));          // top of 'values' contains result, return         return double.parsedouble(df.format(double.parsedouble(string.valueof(values.pop()))));     }      // returns true if 'op2' has higher or same precedence 'op1',     // otherwise returns false.     public static boolean hasprecedence(char op1, char op2)     {         if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-'))             return false;         else             return true;     }      // utility method apply operator 'op' on operands 'a'      // , 'b'. return result.     public static double applyop(char op, double b, double a)     {         switch (op)         {         case '+':             return + b;         case '-':             return - b;         case '*':             return * b;         case '/':             if (b == 0)                 throw new                 unsupportedoperationexception("cannot divide zero");             return / b;         }         return 0;     } } 

there going change in part values getting pushed in 'value' stack.

you have done ! more little change:

        while (i < tokens.length && tokens[i] >= '0' && tokens[i] <= '9') 

needs replaced

       while (i < tokens.length && (character.isdigit(tokens[i]) || tokens[i] == '.'))