需要单元转换器的程序逻辑

时间:2013-08-02 17:37:39

标签: android algorithm logic unit-conversion

这是我的Android应用程序单位转换器。我有三个微调器:单元。 例如。角度,度数和弧度。

我为单位微调器添加了一个监听器。选择单位后,微调器将被填充。用户将在中输入 EditText中的输入,按下Calculate按钮后,TextView将包含答案。

我使用if else实现了这一点。

if unit_spinner is Angle
    if from_spinner is Degree
        if to_spinner is Radian
            return input*0.0174532925 //1 degree = 0.0174532925 rad
        else if to_spinner is Gradian
            return input*1.111111111111111 //1 degree = 1.111111111111111 grad

        ...and so on, the cartesian product of all units

对于多个单位来说这变得非常长。所以你能提出另一种逻辑吗?

3 个答案:

答案 0 :(得分:1)

根据您的变量,您可以使用switch语句。

switch(someintegervariable){
  case SOME_INT_CONSTANT_1:
    /* ... */
    break;
  case SOME_INT_CONSTANT_2:
    /* ... */
    break;
  default:
    /* ... */
}

答案 1 :(得分:1)

现在的方式,对于每个有N个单位的类别,你有两个NxN if语句。

if from_spinner is Degree
    if to_spinner is Degree
        return input
    if to_spinner is Radian
        return input * 0.0174532925199
    if to_spinner is Gradian
        return input * 1.11111111111
if from_spinner is Radian
    if to_spinner is Degree
        return input * 57.2957795131
    if to_spinner is Radian
        return input
    if to_spinner is Gradian
        return input * 63.6619772368
if from_spinner is Gradian
    if to_spinner is Degree
        return input * 0.9
    if to_spinner is Radian
        return input * 0.0157079632679
    if to_spinner is Gradian
        return input

相反,选择一个单元作为输入和输出之间的中介。然后你需要N if个语句从输入转换为中间语,N if语句从中间转换为输出,总共2N。

//we will use degrees as the intermediary unit
intermediary = null
//caluclate intermediary
if from_spinner is Degree
    intermediary = input
if from_spinner is Radian
    intermediary = input * 57.2957795131
if from_spinner is Gradian
    intermediary = input * 0.9

//calculate final
if to_spinner is Degree
    return intermediary
if to_spinner is Radian
    return intermediary / 57.2957795131
if to_spinner is Gradian
    return intermediary / 0.9

当你只有三个单位时,它看起来效率不高,但是对于较大的N值,它可以为你节省很多精力。例如,将这个105行双重嵌套方法与使用中间值的29行等效方法进行比较:

if from_spinner is Millimeter
    if to_spinner is Millimeter
        return input
    if to_spinner is Centimeter
        return input * 0.1
    if to_spinner is Meter
        return input * 0.001
    if to_spinner is Kilometer
        return input * 1e-06
    if to_spinner is Inch
        return input * 0.0393700787402
    if to_spinner is Foot
        return input * 0.00328083989501
    if to_spinner is Mile
        return input * 6.2137273665e-07
if from_spinner is Centimeter
    if to_spinner is Millimeter
        return input * 10.0
    if to_spinner is Centimeter
        return input
    if to_spinner is Meter
        return input * 0.01
    if to_spinner is Kilometer
        return input * 1e-05
    if to_spinner is Inch
        return input * 0.393700787402
    if to_spinner is Foot
        return input * 0.0328083989501
    if to_spinner is Mile
        return input * 6.2137273665e-06
if from_spinner is Meter
    if to_spinner is Millimeter
        return input * 1000.0
    if to_spinner is Centimeter
        return input * 100.0
    if to_spinner is Meter
        return input
    if to_spinner is Kilometer
        return input * 0.001
    if to_spinner is Inch
        return input * 39.3700787402
    if to_spinner is Foot
        return input * 3.28083989501
    if to_spinner is Mile
        return input * 0.00062137273665
if from_spinner is Kilometer
    if to_spinner is Millimeter
        return input * 1000000.0
    if to_spinner is Centimeter
        return input * 100000.0
    if to_spinner is Meter
        return input * 1000.0
    if to_spinner is Kilometer
        return input
    if to_spinner is Inch
        return input * 39370.0787402
    if to_spinner is Foot
        return input * 3280.83989501
    if to_spinner is Mile
        return input * 0.62137273665
if from_spinner is Inch
    if to_spinner is Millimeter
        return input * 25.4
    if to_spinner is Centimeter
        return input * 2.54
    if to_spinner is Meter
        return input * 0.0254
    if to_spinner is Kilometer
        return input * 2.54e-05
    if to_spinner is Inch
        return input
    if to_spinner is Foot
        return input * 0.0833333333333
    if to_spinner is Mile
        return input * 1.57828675109e-05
if from_spinner is Foot
    if to_spinner is Millimeter
        return input * 304.8
    if to_spinner is Centimeter
        return input * 30.48
    if to_spinner is Meter
        return input * 0.3048
    if to_spinner is Kilometer
        return input * 0.0003048
    if to_spinner is Inch
        return input * 12.0
    if to_spinner is Foot
        return input
    if to_spinner is Mile
        return input * 0.000189394410131
if from_spinner is Mile
    if to_spinner is Millimeter
        return input * 1609340.0
    if to_spinner is Centimeter
        return input * 160934.0
    if to_spinner is Meter
        return input * 1609.34
    if to_spinner is Kilometer
        return input * 1.60934
    if to_spinner is Inch
        return input * 63359.8425197
    if to_spinner is Foot
        return input * 5279.98687664
    if to_spinner is Mile
        return input

intermediary = null
if from_spinner is Millimeter
    intermediary = input * 0.001
if from_spinner is Centimeter
    intermediary = input * 0.01
if from_spinner is Meter
    intermediary = input * 1.0
if from_spinner is Kilometer
    intermediary = input * 1000.0
if from_spinner is Inch
    intermediary = input * 0.0254
if from_spinner is Foot
    intermediary = input * 0.3048
if from_spinner is Mile
    intermediary = input * 1609.34
if to_spinner is Millimeter
    return intermediary / 0.001
if to_spinner is Centimeter
    return intermediary / 0.01
if to_spinner is Meter
    return intermediary / 1.0
if to_spinner is Kilometer
    return intermediary / 1000.0
if to_spinner is Inch
    return intermediary / 0.0254
if to_spinner is Foot
    return intermediary / 0.3048
if to_spinner is Mile
    return intermediary / 1609.34

答案 2 :(得分:0)

我建议使用switch/case语句。有了这个,您可以为每种单位类型设置一个案例,然后进行计算。

您还可以为每个计算编写方法。这可能会缩短一点,但可能不会太多。

我只想使用case声明。它相对简单,但如果您需要帮助,我可以为您整理一些代码。请告诉我。

以下是我可能使用的代码:

static double DoCalc()
{
    int uid;
    if unit_roller is Angle
        uid = 1;
    else if unit_roller is Length
        uid = 2;
    //etc...

    switch(uid)
        case 1:
            return AngleCalc(input);
        case 2:
            return LengthCalc(input);
        //etc...
        default;
}

static double AngleCalc(double input)
{
    if (from_spinner == to_spinner)
        return input;
    else if (from_spinner is Degree && to_spinner is radian)
        return input*0.0174532925
    //etc...    
}

我会说,一旦我看到它放在一起,这并不是真的更短,说它确实避免使用嵌套的if语句,并且当你有大量的选项时可能会缩短你的代码