java - 将重复的 if-else 语句修改为更清晰的代码

Question

我正在制作一个计算器 Gui。有什么方法可以使这段代码看起来更干净而不重复行。

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，您可以通过引入接受数字和 DoubleBinaryOperator 的新方法来使其变得非常奇特:

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，您就可以在 switch 语句中找到匹配的运算符，然后使用数字和找到的运算符执行该方法