java - 我的 Java 方法运行不正确导致我的程序编译不正确

Question

我正在编写一个程序，它将提示用户输入整数温度，然后提示用户选择该温度的刻度，华氏度、摄氏度或开尔文。然后它会提示用户从 3 个比例选项中选择他们想要转换成的温度，然后计算转换并显示最终温度。我的程序当前运行但不正确，我已经盯着这个问题好几天了，我不确定为什么它会这样编译。我对编码还很陌生，这是我一直在寻找一些建议的类作业。

封装温度转换器；导入java.util.*;

公共(public)类TempConverter{

public static void main(String[] args) 
{
    double temp1;
    double convertTo;
    double finalTemp;
    double currScale;
    char cont='E';
    char process='P';
    int scale1=0;
    int scale2=0;

    do
    {
        //input
        getTemp();
        scaleOne(scale1);
        toScale(scale2);

        //process


        System.out.print("Your final temp is: " +convert()+toScale(scale2));

    }while(cont=='N');
}   
    private static double getTemp()
    {
         double temp;
         double currentTemp;
         Scanner userIn= new Scanner(System.in);
         System.out.print("Enter a temperature to convert ");
         while(!userIn.hasNextDouble())
         {
          System.out.println("That is not a valid temperature!");
          System.out.print("Re-enter a temperature: ");
          userIn.next();
         }
         temp=userIn.nextDouble();



         return temp;
    }

    private static String scaleOne(int userChoice)
    {

        Scanner userIn= new Scanner(System.in);



        System.out.print("What scale is your temp currently in?"+
                          "1: Fahrenheit" + 
                          "2: Celsius" + 
                          "3: Kelvin" + 
                          "Please enter 1, 2, or 3: ");


        String scale="";

        switch(userChoice)
        {
            case 1:
            {scale= "Fahrenheit";
            break;}

            case 2:
            {scale="Celsius";
            break;}

            case 3:
            {scale="Kelvin";
            break;}

            default:
            {scale= "There was an error in your choice!";}
        }

        return scale;

    }



    private static String toScale(int userChoice)
    {

        Scanner userIn= new Scanner(System.in);



        System.out.print("What scale would you like to convert to?"+
                          "1: Fahrenheit" + 
                          "2: Celsius" + 
                          "3: Kelvin" + 
                          "Please enter 1, 2, or 3: ");


        String scale2="";

        switch(userChoice)
        {
            case 1:
            {scale2= "Fahrenheit";
            break;}

            case 2:
            {scale2="Celsius";
            break;}

            case 3:
            {scale2="Kelvin";
            break;}

            default:
            {scale2= "There was an error in your choice!";}
        }

        return scale2;

    }



    public static double convert()
    {
        double farToKel;
        double farToCel;
        double celToKel;
        double celToFar;
        double kelToFar;
        double kelToCel;
        int scale1=0;
        double converts;

        if(scaleOne(scale1)=="Fahrenheit" && toScale(scale1)=="Kelvin")        
        {
         converts=farToKel=(getTemp()+459.67)* 5/9;
        }

        else if(scaleOne(scale1)=="Fahrenheit" && toScale(scale1)=="Celsius")
        {
         converts=farToCel=(getTemp()-32)*5/9;
        }

        else if(scaleOne(scale1)=="Celsius" && toScale(scale1)=="Kelvin")
        {
         converts=celToKel=getTemp()+273.15;
        }

        else if(scaleOne(scale1)=="Celsius" && toScale(scale1)=="Fahrenheit")
        {
         converts=celToFar=getTemp()*9/5+32;
        }

        else if(scaleOne(scale1)=="Kelvin" && toScale(scale1)=="Fahrenheit")
        {
         converts=kelToFar=getTemp()*9/5-459.67;
        }

        else
        { 
         converts=kelToCel=getTemp()-276.15;
        }

        return converts;
    }








    }

这是它编译的内容

运行:输入温度以转换为 100您当前的温度是多少单位？1: 华氏度2: 摄氏度3: 开尔文请输入1, 2, 或3: 您想要转换成什么单位？1: 华氏度2: 摄氏度3: 开尔文请输入1, 2, 或3: 什么单位您当前的温度是多少？1: 华氏度2: 摄氏度3: 开尔文请输入1, 2, 或3: 您当前的温度是多少度？1: 华氏度2: 摄氏度3: 开尔文请输入1, 2, 或3: 您当前的温度是多少度?1: 华氏度2: 摄氏度3: 开氏度请输入1, 2, 或3: 您当前的温度是多少度？1: 华氏度2: 摄氏度3: 开尔文请输入1, 2, 或3: 您当前的温度是多少度？1: 华氏度2 :摄氏度3:开尔文请输入1、2或3:输入温度以转换2您想转换成什么温标？1:华氏度2:摄氏度3:开尔文请输入1、2或3:您的最终温度是:-274.15您的选择有误!构建成功(总时间:33秒)

这就像是没有以正确的顺序进行编译，或者以正确的顺序获取正确的信息并一遍又一遍地获取某些信息。一旦您输入数字温度，它会提示您输入比例，但不仅仅是一次打印出来，然后您输入第一个比例的选择，它会一直完成，而无需您输入toScale 的选择。

Answer 1

请参阅下面在代码中插入的注释和所做的更正。

public class TempConverter {

    // variables that are needed in more than one method must be declared outside the methods
    // (static since the methods are static; not recommended, but let it be for now)
    static double temp1;
    static String scale1;
    static String scale2;

    public static void main(String[] args) {
        // get rid of unused variables
        char cont = 'E';

        do {
            // input
            // save all inputs to variables
            temp1 = getTemp();
            // Java parameters are pass by value, so to get a value out of the method, use the returned value
            // then you don’t need the parameter
            scale1 = scaleOne();
            scale2 = toScale();

            // process
            // don’t call toScale() again, or the user will be prompted again,
            // instead, use the value already stored in scale2
            System.out.print("Your final temp is: " + convert() + ' ' + scale2);

            // you should set cont to something if you want the loop to repeat

        } while (cont == 'N');
    }

    private static double getTemp() {
        double temp;
        Scanner userIn = new Scanner(System.in);
        System.out.print("Enter a temperature to convert ");
        while (!userIn.hasNextDouble()) {
            System.out.println("That is not a valid temperature!");
            System.out.print("Re-enter a temperature: ");
            userIn.next();
        }
        temp = userIn.nextDouble();

        return temp;
    }

    private static String scaleOne() {

        Scanner userIn = new Scanner(System.in);

        // \n is a line break for readability
        System.out.print("What scale is your temp currently in?" 
                + "\n1: Fahrenheit" 
                + "\n2: Celsius" 
                + "\n3: Kelvin"
                + "\nPlease enter 1, 2, or 3: ");

        // read a value from the user
        int userChoice = userIn.nextInt();

        String scale = "";

        switch (userChoice) {
        case 1: {
            scale = "Fahrenheit";
            break;
        }

        case 2: {
            scale = "Celsius";
            break;
        }

        case 3: {
            scale = "Kelvin";
            break;
        }

        default: {
            scale = "There was an error in your choice!";
        }
        }

        return scale;

    }

    private static String toScale() {

        Scanner userIn = new Scanner(System.in);

        System.out.print("What scale would you like to convert to?" + "1: Fahrenheit" + "2: Celsius" + "3: Kelvin"
                + "Please enter 1, 2, or 3: ");

        String scale2 = "";

        int userChoice = userIn.nextInt();

        // ideally should not accept the same scale as scale1 (from scale)
        switch (userChoice) {
        case 1: {
            scale2 = "Fahrenheit";
            break;
        }

        case 2: {
            scale2 = "Celsius";
            break;
        }

        case 3: {
            scale2 = "Kelvin";
            break;
        }

        default: {
            scale2 = "There was an error in your choice!";
        }
        }

        return scale2;

    }

    public static double convert() {
        // need only one variable for result
        double converts;

        // don’t call methods again, use variables
        // don’t use == for comparing strings, use .equals
        // typo: the second scale corrected to scale2
        if (scale1.equals("Fahrenheit") && scale2.equals("Kelvin")) {
            // don’t call getTemp() again
            converts = (temp1 + 459.67) * 5 / 9;
        }

        else if (scale1.equals("Fahrenheit") && scale2.equals("Celsius")) {
            converts = (temp1 - 32) * 5 / 9;
        }

        else if (scale1.equals("Celsius") && scale2.equals("Kelvin")) {
            converts = temp1 + 273.15;
        }

        else if (scale1.equals("Celsius") && scale2.equals("Fahrenheit")) {
            converts = temp1 * 9 / 5 + 32;
        }

        else if (scale1.equals("Kelvin") && scale2.equals("Fahrenheit")) {
            converts = temp1 * 9 / 5 - 459.67;
        }

        else { // Kelvin to Celsius
            converts = temp1 - 276.15;
        }

        return converts;
    }

}

我相信它现在基本上可以工作了。