中缀表达式求解器中的空堆栈异常

Question

所以我的程序应该采用语法正确的中缀表达式包含整数操作数和来自 GUI 的四个算术运算符 （+ - */） 并显示结果。不幸的是，如果我输入一个简单的表达式，如 3 + 4，程序会抛出一个空堆栈异常。我才刚刚开始使用堆栈，所以如果它简单到我出错，请原谅我。请帮忙！

import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.util.*;

public class inFix extends JFrame
{
   private Container contents;
   private JLabel infixLabel, resultLabel, result;
   private JTextField infixText;
   private JButton evaluate;
   public inFix()
   {
      super("Infix Expresion Evaluator");
      contents = getContentPane();
      contents.setLayout( new FlowLayout() );
      infixLabel = new JLabel("Enter Infix Expression:");
      infixText = new JTextField("", 14);
      evaluate = new JButton("Evaluate");
      resultLabel = new JLabel("Result:");
      result = new JLabel("???");
      contents.add( infixLabel );
      contents.add( infixText );
      contents.add( evaluate );
      contents.add( resultLabel );
      contents.add( result );

      ButtonHandler bh = new ButtonHandler();
      evaluate.addActionListener( bh );
      setSize( 350, 200 );
      setVisible( true );
   }
   private class ButtonHandler implements ActionListener
   {
      public void actionPerformed( ActionEvent ae )
      {
         result.setText(infix(infixText.getText()));
      }
   }

   public String infix(String expression)
   {
      expression=expression.replaceAll("[tn ]", "")+"=";
      String operator = "*/+-";
      int value1, value2;
      char ch;
      StringTokenizer tokenizer = new StringTokenizer(expression, operator, true); 
      Stack<Integer> valueStack = new Stack<Integer>();
      Stack<Character> operatorStack = new Stack<Character>();
      while(tokenizer.hasMoreTokens())
      {
         String token = tokenizer.nextToken();
         if(isInteger(token) == true)
            valueStack.push(Integer.parseInt(token));
         else if(token.charAt(0) == '(')
            operatorStack.push(token.charAt(0));
         else if(token.charAt(0) == ')')
            while(operatorStack.peek() != '(')
            {
               value1 = valueStack.pop();
               value2 = valueStack.pop();
               valueStack.push(solver(value1, value2, operatorStack.pop()));
            }
         else if(token.charAt(0) == '+' || token.charAt(0) == '-' || token.charAt(0) == '*' || token.charAt(0) == '/')
         {
            while(!operatorStack.isEmpty() && precedence(token.charAt(0)) <= precedence(operatorStack.peek()))
            {
               value1 = valueStack.pop();
               value2 = valueStack.pop();
               valueStack.push(solver(value1, value2, token.charAt(0)));
            }
            operatorStack.push(token.charAt(0));
         }
      }
      while(!operatorStack.isEmpty())
      {
         value1 = valueStack.pop();
         value2 = valueStack.pop();
         ch = operatorStack.pop();
         valueStack.push(solver(value1, value2, ch));
      }

      String result = Integer.toString(valueStack.pop());
      return result;     
   }
   public static boolean isInteger(String s)
   {
      try
      { 
         Integer.parseInt(s);
      } 
      catch(NumberFormatException e)
      {
         return false; 
      } 
      catch(NullPointerException e) 
      {
         return false;
      }
      return true;
   }
   public int solver( int value1, int value2, char operator)
   {
      if(operator == '*')
         return value1 * value2;
      else if(operator == '/')
         return value1 / value2;
      else if(operator == '+')
         return value1 + value2;
      else if(operator == '-')
         return value1 - value2;
      else
         return 0;
   }
   public int precedence(char op)
   {
      if(op == '+' || op == '-')
         return 1;
      else if(op == '*' || op == '/')
         return 2;
      else
         return -1;
   }   

   public static void main( String [] args )
   {
      inFix infixsolver = new inFix();
      infixsolver.setDefaultCloseOperation( JFrame.EXIT_ON_CLOSE );
   }
}

Answer 1

假设输入表达式为"3+4"在第 54 行

expression=expression.replaceAll("[tn ]", "")+"=";

表达式变为"3+4="在分词器操作之后，标记为 ["3"，"+"，"4="]，这会在第 65 行生成标记"4="的错误结果

if(isInteger(token) == true)

因此，当您在第 89 行弹出时，值的数量是不够的

     value2 = valueStack.pop();

要解决此问题，请删除第 54 行中的 +"="

expression=expression.replaceAll("[tn ]", "")+"=";