如何在不使用多個返回的情況下返回值

Question

我正在構建一個程序，我們特別不允許在子模塊中使用多個返回。

我想知道如何通過jimArea，ashtynArea和steveArea從子模塊areaCalc，使他們能夠在主要使用。這裏是我所指的子模塊，後面是完整的代碼。

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); 
} 

這是完整的代碼。本來我只是回來了，但事實證明我需要3個領域，所以我不知道如何做到這一點，而不做多次回報。

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; 
} 
} 

任何幫助，非常感謝，謝謝。 約翰

Answer 1

這裏最快的修復可能只是返回一個集合的區域。例如，你可以返回一個List<Integer>而不是單個int，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; 
} 

如果你想獲得比這發燒友，那麼你可以考慮建立一個善意，善意POJO包裝了三倍區的值：

public class AreaClass { 
    private int jimArea; 
    private int steveArea; 
    private int ashtynArea; 

    // constructor, getters and setters 
} 

Answer 2

返回一個對象或數據結構。新程序員傾向於在原語方面考慮太多。你需要將它們組合成對象。

這段代碼是否可以編譯？

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; 
} 

哪裏有jimDepth其餘定義？我在你的main()方法中看到它們，但它們都不在areaCalc()的範圍內。

瞭解數據結構。事情是這樣的：

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; 
    } 
} 

現在你可以有多個不同的每個人：

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)); 

輕鬆添加更多或名稱查找。

Answer 3

的解決方案是使用DTO - 數據傳送對象：類型來包裝乘法對象：

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); 
} 

Answer 4

另外一個選擇是返回一個數組保持所有的值。但我懷疑這會解決你的所有問題，正如duffymo在他的回答中指出的那樣。

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; 
}