Java计算器堆栈

Question

所以我的任务是使用堆栈数据结构来设计一个简单的GUI计算器来执行加法，减法，乘法和除法。计算器窗口应至少有两个面板 - 一个用于显示，另一个用于按钮（0 - 9，。，+， - ，X，/，=，C）。输入表达式后，将显示其后缀或前缀，然后显示结果。

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

public class JCalculator implements ActionListener { 
JFrame jfrm; 
JFrame jfrm2; 
JTextField txt; 
JLabel results; 
String str = ""; 
Stack operands = new Stack(); 
char[] a = new char[0]; 
int used = 0; 



JCalculator(){ 
    jfrm = new JFrame("JCalc"); 
    jfrm.getContentPane().setLayout(new GridLayout(0,1)); 
    jfrm.setSize(210,210); 

    results = new JLabel("",SwingConstants.RIGHT); 
    jfrm.getContentPane().add(results); 
    jfrm.setLocation(400,300); 
    jfrm.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); 

    JButton[] calbut= new JButton[14]; 
    String []but = {"7","8","9","/","4","5","6","*","1","2","3","-","0","C" 
    }; 
    JButton equal = new JButton("="); 
    JButton add = new JButton("+"); 

    JPanel jbutton = new JPanel(); 
    jbutton.setLayout(new GridLayout(4,3)); 
    for(int i=0; i< but.length;i++){ 
     jbutton.add(calbut[i] = new JButton(but[i])); 

     calbut[i].addActionListener(this); 
    } 
    jbutton.add(equal); 
    jbutton.add(add); 

    equal.addActionListener(this); 
    add.addActionListener(this); 

    jfrm.getContentPane().add(jbutton); 


    jfrm.setVisible(true); 
} 
public void actionPerformed(ActionEvent ae){ 
    char[] a = new char[0]; 
    int used = 0; 
    if(used == a.length){ 
     char[] newa= new char[a.length + 1]; 
     for(int i = 0; i <used; i++) newa[i]= a[i]; 
     a= newa; 
    } 

    if(ae.getActionCommand().equals("1")){ 
     str = results.getText(); 
     results.setText(str + "1"); 
     operands.push(1); 
    } 
    if(ae.getActionCommand().equals("2")){ 
     str = results.getText(); 
     results.setText(str + "2"); 
     operands.push(2); 
    } 
    if(ae.getActionCommand().equals("3")){ 
     str = results.getText(); 
     results.setText(str + "3"); 
     operands.push(3); 
    } 
    if(ae.getActionCommand().equals("4")){ 
     str = results.getText(); 
     results.setText(str + "4"); 
     operands.push(4); 
    } 
    if(ae.getActionCommand().equals("5")){ 
     str = results.getText(); 
     results.setText(str + "5"); 
     operands.push(5); 
    } 
    if(ae.getActionCommand().equals("6")){ 
     str = results.getText(); 
     results.setText(str + "6"); 
     operands.push(6); 
    } 
    if(ae.getActionCommand().equals("7")){ 
     str = results.getText(); 
     results.setText(str + "7"); 
     operands.push(7); 
    } 
    if(ae.getActionCommand().equals("8")){ 
     str = results.getText(); 
     results.setText(str + "8"); 
     operands.push(8); 
    } 
    if(ae.getActionCommand().equals("9")){ 
     str = results.getText(); 
     results.setText(str + "9"); 
     operands.push(9); 
    } 
    if (ae.getActionCommand().equals("0")) { 
     str = results.getText(); 
     results.setText(str + "0"); 
     operands.push(0); 
    } 
    if(ae.getActionCommand().equals("+")){ 
     str = results.getText(); 
     double operand = Double.parseDouble(str); 
     operands.push(operand); 
     results.setText(""); 
     a[used]= '+'; 
     used++; 

    } 
    if(ae.getActionCommand().equals("-")){ 
     str = results.getText(); 
     double operand = Double.parseDouble(str); 
     operands.push(operand); 
     results.setText(""); 
     a[used]= '+'; 
     used++; 
    } 
    if(ae.getActionCommand().equals("/")){ 
     str = results.getText(); 
     double operand = Double.parseDouble(str); 
     operands.push(operand); 
     results.setText(""); 
     a[used]= '+'; 
     used++; 
    } 
    if(ae.getActionCommand().equals("*")){ 
     str = results.getText(); 
     double operand = Double.parseDouble(str); 
     operands.push(operand); 
     results.setText(""); 
     a[used]= '*'; 
     used++; 
    } 
    if(ae.getActionCommand().equals("=")){ 

     for(int i = 0; i< used; i++){ 
      performBinaryOp(a[i]); 
     } 

     str = String.valueOf(operands.pop()); 
     results.setText(str); 
    } 


} 
public void performBinaryOp(char nextOperation) { 
    double leftOperand, rightOperand; 
    Double result = new Double(0); 
    rightOperand = (Double) operands.pop(); 
    leftOperand = (Double) operands.pop(); 
    switch (nextOperation) { 
     case '+': 
      result = new Double(leftOperand + rightOperand); 
      break; 
     case '-': 
      result = new Double(leftOperand - rightOperand); 
      break; 
     case '*': 
      result = new Double(leftOperand * rightOperand); 
      break; 
     case '/': 
      result = new Double(leftOperand/rightOperand); 
      break; 
    } 
    operands.push(result); 
} 

public static void main(String[] args) 
{ 
    SwingUtilities.invokeLater(new Runnable() { 
     public void run() { 
      new JCalculator();     
     }    
    });   
} 
} 

，我碰上是当按下按钮等于，例如按下2 + 3 =它返回第一个值2.0误差然后我收到一条错误消息：

Exception in thread "AWT-EventQueue-0" java.util.EmptyStackException 
at java.util.Stack.peek(Unknown Source) 
at java.util.Stack.pop(Unknown Source) 
at JCalculator.actionPerformed(JCalculator.java:152) 
at javax.swing.AbstractButton.fireActionPerformed(Unknown Source) 
at javax.swing.AbstractButton$Handler.actionPerformed(Unknown Source) 
at javax.swing.DefaultButtonModel.fireActionPerformed(Unknown Source) 
at javax.swing.DefaultButtonModel.setPressed(Unknown Source) 
at javax.swing.plaf.basic.BasicButtonListener.mouseReleased(Unknown Source) 
at java.awt.Component.processMouseEvent(Unknown Source) 
at javax.swing.JComponent.processMouseEvent(Unknown Source) 
at java.awt.Component.processEvent(Unknown Source) 
at java.awt.Container.processEvent(Unknown Source) 
at java.awt.Component.dispatchEventImpl(Unknown Source) 
at java.awt.Container.dispatchEventImpl(Unknown Source) 
at java.awt.Component.dispatchEvent(Unknown Source) 
at java.awt.LightweightDispatcher.retargetMouseEvent(Unknown Source) 
at java.awt.LightweightDispatcher.processMouseEvent(Unknown Source) 
at java.awt.LightweightDispatcher.dispatchEvent(Unknown Source) 
at java.awt.Container.dispatchEventImpl(Unknown Source) 
at java.awt.Window.dispatchEventImpl(Unknown Source) 
at java.awt.Component.dispatchEvent(Unknown Source) 
at java.awt.EventQueue.dispatchEventImpl(Unknown Source) 
at java.awt.EventQueue.access$000(Unknown Source) 
at java.awt.EventQueue$1.run(Unknown Source) 
at java.awt.EventQueue$1.run(Unknown Source) 
at java.security.AccessController.doPrivileged(Native Method) 
at java.security.AccessControlContext$1.doIntersectionPrivilege(Unknown Source) 
at java.security.AccessControlContext$1.doIntersectionPrivilege(Unknown Source) 
at java.awt.EventQueue$2.run(Unknown Source) 
at java.awt.EventQueue$2.run(Unknown Source) 
at java.security.AccessController.doPrivileged(Native Method) 
at java.security.AccessControlContext$1.doIntersectionPrivilege(Unknown Source) 
at java.awt.EventQueue.dispatchEvent(Unknown Source) 
at java.awt.EventDispatchThread.pumpOneEventForFilters(Unknown Source) 
at java.awt.EventDispatchThread.pumpEventsForFilter(Unknown Source) 
at java.awt.EventDispatchThread.pumpEventsForHierarchy(Unknown Source) 
at java.awt.EventDispatchThread.pumpEvents(Unknown Source) 
at java.awt.EventDispatchThread.pumpEvents(Unknown Source) 
at java.awt.EventDispatchThread.run(Unknown Source) 

我很确定我的问题是我有一个空的堆栈，但我不知道代码错在哪里或如何解决它，所以任何帮助将不胜感激。

Answer 1

我建议您将Swing和UI放在一边，直到您的计算器正常工作。创建一个单独的Calculator类，可以接受文本字符串并对它们进行评估。一旦你有这个工作，然后添加一个用户界面。

是的，你已经有了一个栈的问题，就如同例外告诉你：

at JCalculator.actionPerformed(JCalculator.java:152) 

在文本编辑器打开你的JCalculator，把行数显示，并转到行152这是你的问题在哪里。

我不愿意通读您发布的代码，但我会说与调试器快速旋转应该很快就清除它。

Answer 2

只需在您需要用计算器帮助的情况下，这里是在Java中我的第一个项目

https://github.com/orefalo/ExpressionEvaluator

Answer 3

你actionPerformed()方法shadows一些类变量;请注意，您已重新声明char [] a和int used，因此您的代码几乎肯定不会按照您的预期运行。我没有遵循这个逻辑结论。

public class JCalculator implements ActionListener { 
JFrame jfrm; 
JFrame jfrm2; 
JTextField txt; 
JLabel results; 
String str = ""; 
Stack operands = new Stack(); 
char[] a = new char[0]; 
int used = 0; 
/* ... */ 
public void actionPerformed(ActionEvent ae){ 
    char[] a = new char[0]; 
    int used = 0; 
    if(used == a.length){ 
     char[] newa= new char[a.length + 1]; 
     for(int i = 0; i <used; i++) newa[i]= a[i]; 
     a= newa; 
    } 

这部分代码让我觉得，你不接受两个数字：

if(ae.getActionCommand().equals("1")){ 
    str = results.getText(); 
    results.setText(str + "1"); 
    operands.push(1); 
} 

的operands.push(1)可能须推动数正在生成，而不仅仅是最最近的数字。 （当然，如果你的意图是仅支持单位数字，这看起来没问题。）

我建议试图用单个函数替换那些十个小的if语句。在这里花时间弄清楚一个函数是非常值得的，它将大大简化将来代码的维护，并且技巧是至关重要的。 （我想建议你考虑你的整齐JButton初始化早些时候 - 并尝试在你的替换功能使用数组它可能不是最好的，但它应该是好的。）

if(ae.getActionCommand().equals("/")){ 
    str = results.getText(); 
    double operand = Double.parseDouble(str); 
    operands.push(operand); 
    results.setText(""); 
    a[used]= '+'; 
    used++; 
} 

你会注意到你的a[used]被分配了错误的符号。一旦你修复了你的变量shadowing问题，这个将成为一个问题。尝试找出一种方法，用另一种单一功能取代这四个if块 - 再一次，当向计算器添加新操作符时，它会使这种错误变得更加困难，并且会修复运算符应用代码中的错误更容易，因为你只需要修复一个位置的错误。

我不能强调足够强调将代码分解成“最小合理”功能的有用程度。您可以测试小函数比大函数更容易，并且将来您的代码将更易于阅读。