Question

我用Java制作了这个计算器程序。这仅在一次计算两个数字时才有效。这意味着要获得1 + 2 + 3的总和，你必须这样：按1 按+ 按2 按= 按+ 按3 按=

并将其计算为6。

但我想对此进行编程，以便我可以通过以下方式获得答案：按1 按+ 按2 按+ 按3 按=

但这给出了答案5 !!!如何对其进行编码以使其像普通计算器一样工作？

这是我的代码：

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

public class cal1 extends JFrame {

double op1 = 0d, op2 = 0d;
double result = 0d;
char action;
boolean b = false;
boolean pressequal = false;

public cal1() {
    makeUI();
}

private void makeUI() {
    setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
    setSize(400, 400);

    b0 = new JButton("0");
    b1 = new JButton("1");
    b2 = new JButton("2");
    b3 = new JButton("3");
    b4 = new JButton("4");
    b5 = new JButton("5");
    b6 = new JButton("6");
    b7 = new JButton("7");
    b8 = new JButton("8");
    b9 = new JButton("9");
    bDot = new JButton(".");
    bMul = new JButton("*");
    bDiv = new JButton("/");
    bPlus = new JButton("+");
    bMinus = new JButton("-");
    bEq = new JButton("=");

    t = new JTextField(12);
    t.setFont(new Font("Tahoma", Font.PLAIN, 24));
    t.setHorizontalAlignment(JTextField.RIGHT);

    numpad = new JPanel();
    display = new JPanel();

    numpad.add(b7);
    numpad.add(b8);
    numpad.add(b9);
    numpad.add(bMul);
    numpad.add(b4);
    numpad.add(b5);
    numpad.add(b6);
    numpad.add(bDiv);
    numpad.add(b1);
    numpad.add(b2);
    numpad.add(b3);
    numpad.add(bMinus);
    numpad.add(bDot);
    numpad.add(b0);
    numpad.add(bEq);
    numpad.add(bPlus);

    numpad.setLayout(new GridLayout(4, 5, 5, 4));

    display.add(t);
    add(display, BorderLayout.NORTH);
    add(numpad, BorderLayout.CENTER);

    t.addKeyListener(new KeyAdapter() {
        @Override
        public void keyTyped(KeyEvent e) {
            typeOnt(e);
        }
    });

    b0.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b0pressed(e);
        }
    });

    b1.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b1pressed(e);
        }
    });

    b2.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b2pressed(e);
        }
    });

    b3.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b3pressed(e);
        }
    });

    b4.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b4pressed(e);
        }
    });

    b5.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b5pressed(e);
        }
    });

    b6.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b6pressed(e);
        }
    });

    b7.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b7pressed(e);
        }
    });

    b8.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b8pressed(e);
        }
    });

    b9.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            b9pressed(e);
        }
    });

    bDot.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            bDotpressed(e);
        }
    });

    bPlus.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            bPlusPressed(e);
        }
    });

    bMinus.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            bMinusPressed(e);
        }
    });

    bMul.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            bMulPressed(e);
        }
    });

    bDiv.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            bDivPressed(e);
        }
    });

    bEq.addActionListener(new ActionListener() {
        @Override
        public void actionPerformed(ActionEvent e) {
            bEqpressed(e);
        }
    });

}

void typeOnt(KeyEvent e) {
    e.consume();
}

void b0pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "0");
    } else {
        t.setText(t.getText() + "0");
    }
}

void b1pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "1");
    } else {
        t.setText(t.getText() + "1");
    }
}

void b2pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "2");
    } else {
        t.setText(t.getText() + "2");
    }
}

void b3pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "3");
    } else {
        t.setText(t.getText() + "3");
    }
}

void b4pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "4");
    } else {
        t.setText(t.getText() + "4");
    }
}

void b5pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "5");
    } else {
        t.setText(t.getText() + "5");
    }
}

void b6pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "6");
    } else {
        t.setText(t.getText() + "6");
    }
}

void b7pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "7");
    } else {
        t.setText(t.getText() + "7");
    }
}

void b8pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "8");
    } else {
        t.setText(t.getText() + "8");
    }
}

void b9pressed(ActionEvent e) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + "9");
    } else {
        t.setText(t.getText() + "9");
    }
}

void bDotpressed(ActionEvent e) {
    if (!t.getText().contains(".")) {
        if (b) {
            t.setText(null);
            b = false;
            t.setText(t.getText() + "0.");
        } else if (t.getText().isEmpty()) {
            t.setText("0.");
        } else {
            t.setText(t.getText() + ".");
        }
    }
}

void bPlusPressed(ActionEvent e) {
    b = true;
    action = '+';
    op1 = Double.parseDouble(t.getText());

}

void bMinusPressed(ActionEvent e) {
    b = true;
    action = '-';
    op1 = Double.parseDouble(t.getText());
}

void bMulPressed(ActionEvent e) {
    b = true;
    action = '*';
    op1 = Double.parseDouble(t.getText());
}

void bDivPressed(ActionEvent e) {
    b = true;
    action = '/';
    op1 = Double.parseDouble(t.getText());
}

void bEqpressed(ActionEvent e) {
    op2 = Double.parseDouble(t.getText());
    doCal();
}

void doCal() {
    switch (action) {
        case '+': result = op1 + op2; break;
        case '-': result = op1 - op2; break;
        case '*': result = op1 * op2; break;
        case '/': result = op1 / op2; break;
    }      
    t.setText(String.valueOf(result));
}

public static void main(String[] args) {
    new cal1().setVisible(true);
}

JButton b0;
JButton b1;
JButton b2;
JButton b3;
JButton b4;
JButton b5;
JButton b6;
JButton b7;
JButton b8;
JButton b9;
JButton bDot;
JButton bPlus;
JButton bMinus;
JButton bMul;
JButton bDiv;
JButton bEq;
JPanel display;
JPanel numpad;
JTextField t;
}

Answer 1

Java类中的第一个应该以首字母UpperCase Java Code Conventions

开头

其次，而不是为每个数字创建一个方法

void bpressed(ActionEvent e, Integer number) {
    if (b) {
        t.setText(null);
        b = false;
        t.setText(t.getText() + number);
    } else {
        t.setText(t.getText() + number);
    }
}

使用声明性名称，b是什么？它不直观，类名应该是计算器或类似的东西，代码必须是人类可读的。

与您的问题相关，当某些操作正在进行时，您可以拥有partialResult。例如：

你宣布

private double partialResult=0D;

和类似的方法

private void calculatePartialResult(Integer numberSelected){
      //choose depends on action
      partialResult op= numberSelected;  (where op in + - * /)    
  }

Answer 2

以字符串形式获取计算并使用ScriptEngine：

ScriptEngine engine = new ScriptEngineManager().getEngineByName("JavaScript");
t.setText(engine.eval(calculation_String));

Answer 3

当你做1 + 2 + 3当前代码时：

User tap on '1' '+' 
op1 = 1.0, action = '+' 

User tap on '2' '+'
op1 = 2.0, action = '+'

User tap on '3' '='
op1 = 2.0, op2 = 3.0, action = '+'

doCal() displays the result of op1 + op2 which is 5

我认为您应该使用op1来存储以前计算的值，以获得您期望的行为。

我会这样做。

仅在按下相等时调用doCal()，而不是每次按下操作符时在操作员操作处理程序的开头调用它。将值存储在op2内，以便进行计算：

void bPlusPressed(ActionEvent e) {
    op2 = Double.parseDouble(t.getText());
    doCal();
    b = true;
    action = '+';
}

（对其他运营商应用相同的更改）。

在doCal()内，将结果存储到op1而不是显示它：

void doCal() {
    // No need to use a class attribute for result, prefer a local variable.
    final double result;
    switch (action) {
        case '+': result = op1 + op2; break;
        case '-': result = op1 - op2; break;
        case '*': result = op1 * op2; break;
        case '/': result = op1 / op2; break;
        // When no action is set, simply copy op2 into op1
        default: result = op2;
    }
    op1 = result;
}

然后，更改bEqPressed()以显示op1并重置action的值：

void bEqpressed(ActionEvent e) {
    op2 = Double.parseDouble(t.getText());
    doCal();
    t.setText(String.valueOf(result));
    action = '\0';
}

这应该可以解决问题。

Java简单计算器

3 个答案: