函数返回值始终为零（0.0）

Question

我创建了可转换单位的移动应用，

主Java代码中的一个函数，它接收输入的数字和需要从中转换的单位以及需要转换成的单位，之后进行计算并返回值，但是，问题是函数总是返回值0.0

此功能的以下代码：

   public double areaConvert(double originalNumber, String originalUnit, String newUnit)
    { //Begin convertArea
        //Make two doubles, one that holds the original and one that will be redefined where needed

        //Make two strings, capturing the units fed to the method
        String originalU = originalUnit.toLowerCase();
        String newU = newUnit.toLowerCase();
        double num1 = originalNumber;
        double num2 = 0.0;


        switch(originalU) {
            //Begin unit conversions
            case "12":
                switch (newU) {
                    case "12":
                        num2 = originalNumber;
                        break;
                    case "14":
                        num2 = (num1 * 12) / 14;
                        break;
                    case "18":
                        num2 = (num1 * 12) / 18;
                        break;
                    case "21":
                        num2 = (num1 * 12) / 21;
                        break;
                    case "22":
                        num2 = (num1 *12) / 22;
                        break;
                    case "24":
                        num2 = (num1 *12) /24;
                        break;

                }
                break;
            case "14":
                switch (newU) {
                    case "12":
                        num2  = (num1 * 14) /12;
                        break;
                    case "14":
                        num2 = num1;
                        break;
                    case "18":
                        num2 = (num1 * 14) / 18;
                        break;
                    case "21":
                        num2 = (num1 * 14)  /21;
                        break;
                    case "22":
                        num2 = (num1 *14) / 22;
                        break;
                    case "24":
                        num2 = (num1 *14) / 24;
                        break;
                }
                break;
            case "18":
                switch (newU) {
                    case "12":
                        num2 = (num1 * 18) /12;
                        break;
                    case "14":
                        num2 = (num1 * 18) / 14;
                        break;
                    case "18":
                        num2 = num1 ;
                        break;
                    case "21":
                        num2 = (num1 * 18) / 21;
                        break;
                    case "22":
                        num2 = (num1  * 18) / 22;
                        break;
                    case "24":
                        num2 = (num1 * 18) / 24;
                        break;
                }
                break;
            case "21":
                switch (newU) {
                    case "12":
                        num2 = (num1 * 21) /12;
                        break;
                    case "14":
                        num2 = (num1 * 21) / 14;
                        break;
                    case "18":
                        num2 = (num1 * 21) / 18;
                        break;
                    case "21":
                        num2 = num1 ;
                        break;
                    case "22":
                        num2 = (num1 * 21) / 22;
                        break;
                    case "24":
                        num2 = (num1 * 21) / 24;
                        break;
                }
                break;
            case "22":
                switch (newU) {
                    case "12":
                        num2 = (num1 * 22) /12;
                        break;
                    case "14":
                        num2 = (num1 * 22) / 14;
                        break;
                    case "18":
                        num2 = (num1  * 22) / 18;
                        break;
                    case "21":
                        num2 = (num1 * 22) / 21;
                        break;
                    case "22":
                        num2 = num1 ;
                        break;
                    case "24":
                        num2 = (num1 * 22) / 24;
                        break;
                }
                break;
            case "24":
                switch (newU) {
                    case "12":
                        num2 = (num1 * 24) /12;
                        break;
                    case "14":
                        num2 = (num1 * 24) / 14;
                        break;
                    case "18":
                        num2 = (num1 * 24) / 18;
                        break;
                    case "21":
                        num2 = (num1 * 24) / 21;
                        break;
                    case "22":
                        num2 = (num1 * 24) / 22;
                        break;
                    case "24":
                        num2 = num1 ;
                        break;
                }
                break;

        } //End conversion table
        return num2;
        //Return the resulting number from the conversion table above

   } //End convertArea

} //End converter class

Answer 1

它可能返回零，因为您没有执行任何将其设置为非零的case子句。

我建议对'default：'案例进行编码，以检测不适当的输入。也许只是打印出“无法到达这里”以及您所打开的内容的值。

Answer 2

在我看来，您需要的是

num2 = num1 * Double.parseDouble(originalU) / Double.parseDouble(newU);

Answer 3

如果将3.00，“ 12”，“ 12”传递给该函数，它将返回3.00，恰好是您所期望的。它按照预期的方式工作。我不明白您的意思是返回0.0吗？主要问题是测试每个数字的功能有多大？设计真的很糟糕。您也不会处理传递给此方法的意外值，这将破坏代码并使应用程序崩溃。当您传递数字时，我看不到toLowerCase（）的原因。