Question

我正在制作一个计算器Gui。有没有办法让这个代码看起来更干净，没有重复行。

double num1, num2, ans, doub;
num1 = Double.parseDouble(FirstInput.getText());
num2 = Double.parseDouble(SecondInput.getText());
oper = (Operator.getText());

if(oper.equals("+")){
    ans = num1 + num2;
    doub = Math.round(ans * 100000.0) / 100000.0;
    if(doub == (int) ans){
        Answer.setText(Integer.toString((int) ans));
    } else{
        Answer.setText(Double.toString(doub));
    }
}
if(oper.equals("-")){
    ans = num1 - num2;
    doub = Math.round(ans * 100000.0) / 100000.0;
    if(doub == (int) ans){
        Answer.setText(Integer.toString((int) ans));
    } else{
        Answer.setText(Double.toString(doub));
    }
}
if(oper.equals("/")){
    ans = num1 / num2;
    doub = Math.round(ans * 100000.0) / 100000.0;
    if(doub == (int) ans){
        Answer.setText(Integer.toString((int) ans));
    } else{
        Answer.setText(Double.toString(doub));
    }
}
if(oper.equals("x")){
    ans = num1 * num2;
    doub = Math.round(ans * 100000.0) / 100000.0;
    if(doub == (int) ans){
        Answer.setText(Integer.toString((int) ans));
    } else{
        Answer.setText(Double.toString(doub));
    }
}

Answer 1

使用java-8，你可以通过引入一个接受数字和DoubleBinaryOperator的新方法来使它变得非常有趣：

public void setAnswerText(double num1, double num2, DoubleBinaryOperator operator){
    final double ans = operator.applyAsDouble(num1, num2);
    final double doub = Math.round(ans * 100000.0) / 100000.0;
    if(doub == (int) ans){
        Answer.setText(Integer.toString((int) ans));
    } else{
        Answer.setText(Double.toString(doub));
    }
}

并使用switch语句：

final double num1 = Double.parseDouble(FirstInput.getText());
final double num2 = Double.parseDouble(SecondInput.getText());
final String oper = (Operator.getText());
final DoubleBinaryOperator operator;
switch(oper){
    case "+": 
        operator = (a, b) -> a+b;
        break;
    case "-": 
        operator = (a, b) -> a-b;
        break;
    case "/": 
        operator = (a, b) -> a/b;
        break;
    case "x": 
        operator = (a, b) -> a*b;
        break;
    default:
        throw new UnsupportedOperationException();
}

setAnswerText(num1, num2, operator);

这样，你在switch-statment中找到了匹配的运算符，然后用数字和找到的运算符执行方法

Answer 2

您可以将准备答案的部分提取到单独的方法中，这样您就不需要多次编写该代码，例如

private static void prepAnswer(double ans){
    double doub = Math.round(ans * 100000.0) / 100000.0;
    if(doub == (int) ans){
        Answer.setText(Integer.toString((int) ans));
    } else{
        Answer.setText(Double.toString(doub));
    }
}

此外，您可以使用switch-case语句。这看起来像这样：

    String oper = (Operator.getText());
    switch(oper) {
        case "+": prepAnswer(num1 + num2); break;
        case "-": prepAnswer(num1 - num2); break;
        case "/": prepAnswer(num1 / num2); break;
        case "x": prepAnswer(num1 * num2); break;    
        default: throw new UnsupportedOperationException();
    }

Answer 3

只是巩固

    double num1, num2, ans, doub;
    num1 = Double.parseDouble(FirstInput.getText());
    num2 = Double.parseDouble(SecondInput.getText());
    oper = (Operator.getText());

    switch(oper){
    case "+" :  ans = num1 + num2; break;
    case "-" :  ans = num1 - num2; break;
    case "/" :  ans = num1 / num2; break;
    case "*" :  ans = num1 * num2; break;
    }
    setAnswer(ans);
}

    /**
     * @param ans
     */
    private static void setAnswer(double ans) {
        double doub;
        doub = Math.round(ans * 100000.0) / 100000.0;
        if (doub == (int) ans) {
            Answer.setText(Integer.toString((int) ans));
        } else {
            Answer.setText(Double.toString(doub));
        }
    }

Answer 4

我喜欢Lino的回答，使用DoubleBinaryOperator。（+ 1）。大多数ifs或switch可以被多态替换。没有ifs / switch的功能更易于阅读/测试和维护。受到谷歌谈话的影响很大https://www.youtube.com/watch?v=4F72VULWFvc 您也可以使用此ENUM版本。你需要JDK 8

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将重复的if-else语句修改为更干净的代码

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