Java 如何在不使用多个返回的情况下返回值

我正在构建一个程序，我们特别不允许在子模块中使用多个返回

我想知道如何从子模块

areaCalc

传递

jimArea

、

ashtynArea

和

steveArea

，以便它们可以在main中使用。这是我指的子模块，后面是完整的代码

public static int areaCalc(double depth, double width, double length)
{
 int jimArea = (int)(jimDepth * jimWidth * jimLength);
 int steveArea = (int)(steveDepth * steveWidth * steveLength);
 int ashtynArea = (int)(ashtynDepth * ashtynWidth * ashtynLength);
}

这是完整的代码。最初我只需要返回区域，但结果我需要3个区域，所以我不知道如何在不进行多次返回的情况下完成

import java.util.*;
public class PoolStorageCalculation
{
public static void main(String [] args)
{
 Scanner sc = new Scanner(System.in);
 System.out.println("Please enter the Depth of Steve's Pool in Metres.");
 double steveDepth = sc.nextDouble();
 System.out.println("Please enter the Width of Steve's Pool in Metres.");
 double steveWidth = sc.nextDouble();
 System.out.println("Please enter the Length of Steve's Pool in Metres.");
 double steveLength = sc.nextDouble();
 System.out.println("Please enter the Depth of Jim's Pool in Metres.");
 double jimDepth = sc.nextDouble()
 System.out.println("Please enter the Width of Jim's Pool in Metres.");
 double jimWidth = sc.nextDouble();
 System.out.println("Please enter the Length of Jim's Pool in Metres.");
 double jimLength = sc.nextDouble;
 System.out.println("Please enter the Depth of Ashtyn's Pool in Metres.");
 double ashtynDepth = sc.nextDouble();
 System.out.println("Please enter the Width of Ashtyn's Pool in Metres.");
 double ashtynWidth = sc.nextDouble();
 Systemm.out.println("Please enter the Length of Ashtyn's Pool in Metres.");
 double ashtynLength = sc.nextDouble();
 int area = areaCalc(steveDepth,steveWidth,steveLength,jimDepth,jimWidth,jimLength,ashtynDepth,ashtynLength,ashtynWidth);
 int numRays = rayCalc(steveArea);
 int numSharks = sharkCalc(jimArea);
 int numTurtles = turtleCalc(ashtynArea);
 System.out.println("Steve can store " + numRays + " Sting Rays in his " + steveArea + " Metres Cubed Pool.");
 System.out.println("Jim can store " + numSharks + " Sharks in his " + jimArea + " Metres Cubed Pool.");
 System.out.println("Ashtyn can store " + numTurtles + " Turtles in her " + ashtynArea + " Metres Cubed Pool.");
}
public static int areaCalc(double depth, double width, double length)
{
 int jimArea = (int)(jimDepth * jimWidth * jimLength);
 int steveArea = (int)(steveDepth * steveWidth * steveLength);
 int ashtynArea = (int)(ashtynDepth * ashtynWidth * ashtynLength);

 return area;
}
public static int rayCalc(int steveArea)
{
 int numRays = (int)(steveArea * 0.5);
 return numRays;
}
public static int sharkCalc(int jimArea)
{
 int numSharks = (int)(jimArea * 0.1);
 return numSharks;
}
public static int turtleCalc(int ashtynArea)
{
 int numTurtles = (int)(ashtynArea * 1.2);
 return numTurtles;
}
}

非常感谢您的帮助，谢谢。

---

John

这里最快的修复方法可能就是返回一组区域。例如，您可以返回一个

列表

，而不是单个

int

，例如：

public static List<Integer> areaCalc(double depth, double width, double length)
{
    // Note: you don't actually use the method parameters in the assignments below.
    // This is probably a typo.
    int jimArea = (int)(jimDepth * jimWidth * jimLength);
    int steveArea = (int)(steveDepth * steveWidth * steveLength);
    int ashtynArea = (int)(ashtynDepth * ashtynWidth * ashtynLength);

    List<Integer> areaList = new ArrayList<>();
    areaList.add(jimArea);
    areaList.add(steveArea);
    areaList.add(ashtynArea);

    return areaList;
}

---

返回对象或数据结构。新程序员倾向于在原语方面考虑太多。您需要将它们组合成对象

这段代码甚至可以编译吗

public static int areaCalc(double depth, double width, double length)
{
 int jimArea = (int)(jimDepth * jimWidth * jimLength);
 int steveArea = (int)(steveDepth * steveWidth * steveLength);
 int ashtynArea = (int)(ashtynDepth * ashtynWidth * ashtynLength);

 return area;
}

在哪里定义了

jimDepth

，其余的定义在哪里？我在

main（）

方法中看到了它们，但它们都不在

areaCalc（）的范围内

了解数据结构。大概是这样的：

public class Cube {
    private final double length;
    private final double width;
    private final double depth;

    public Cube(double l, double w, double d) {
        // Question: Do negative or zero dimensions make physical sense?  
        // What would you do about them here?  
        // Hint: Programming by contract and preconditions.
        this.length = l;
        this.width = w;
        this.depth = d;
    }

    public double volume() {
        return this.length*this.width*this.depth;
    }
}

现在，您可以为每个人设置不同的集合：

Map<String, Cube> volumes = new HashMap<>();
volumes.put("jim", new Cube(1, 2, 3));
volumes.put("steve", new Cube(4, 5, 6));
volumes.put("ashtyn", new Cube(7, 8, 9));

Map volumes=newhashmap（）；
体积。放置（“jim”，新立方体（1,2,3））；
卷。放（“史蒂夫”，新立方体（4，5，6））；
体积。放置（“ashtyn”，新立方体（7,8,9））；

易于添加更多或按名称查找。
解决方案是使用dto-数据传输对象：键入以包装多个对象：

public static class AreaDto {
    private int jimArea;
    private int steveArea;
    private int ashtynArea;

    public AreaDto(int jimArea, int steveArea, int ashtynArea) {
        this.jimArea = jimArea;
        this.steveArea = steveArea;
        this.ashtynArea = ashtynArea;
    }

    public int getJimArea() {
        return this.jimArea;
    }

    public int getSteveArea() {
        return this.steveArea;
    }

    public int getAshtynArea() {
        return this.ashtynArea;
    }
}

public static AreaDto areaCalc(int steveDepth, int steveWidth, int steveLength, int jimDepth, int jimWidth, int jimLength, int ashtynDepth, int ashtynLength, int ashtynWidth)
{
    int jimArea = (int)(jimDepth * jimWidth * jimLength);
    int steveArea = (int)(steveDepth * steveWidth * steveLength);
    int ashtynArea = (int)(ashtynDepth * ashtynWidth * ashtynLength);

    return new AreaDto(jimArea, steveArea, ashtynArea);
}

另一个选项是返回一个包含所有值的数组。但我怀疑这能解决你所有的问题，正如达菲莫在回答中指出的那样

public static int[] areaCalc(double depth, double width, double length)
{
 int jimArea = (int)(jimDepth * jimWidth * jimLength);
 int steveArea = (int)(steveDepth * steveWidth * steveLength);
 int ashtynArea = (int)(ashtynDepth * ashtynWidth * ashtynLength);
 int[] area = {jimArea,steveArea,ashtynArea};

 return area;
}

可能值得考虑将面积计算方法更改为仅计算一个面积，并使用不同的输入调用三次。然后，您可以轻松捕获调用方法中计算出的三个区域。@duffymo我知道面向对象，但我们的讲师还没有教给全班，因此我们打算不使用它。Map Data Structure呢？方法areaCalc的参数没有使用。DTO是J2EE反模式。duffymo，对不起，我不同意这里的DTO是反模式的。一般来说，它可能是，但不是在这里。如果需要，可以将其命名为“AreaResult”，并删除“DTO”。但它解决了这个问题。
public static int[] areaCalc(double depth, double width, double length)
{
 int jimArea = (int)(jimDepth * jimWidth * jimLength);
 int steveArea = (int)(steveDepth * steveWidth * steveLength);
 int ashtynArea = (int)(ashtynDepth * ashtynWidth * ashtynLength);
 int[] area = {jimArea,steveArea,ashtynArea};

 return area;
}