问题

这个Java代码：

public class XYZ {   
    public static void main(){  
        int toyNumber = 5;   
        XYZ temp = new XYZ();  
        temp.play(toyNumber);  
        System.out.println("Toy number in main " + toyNumber);  
    }

    void play(int toyNumber){  
        System.out.println("Toy number in play " + toyNumber);   
        toyNumber++;  
        System.out.println("Toy number in play after increement " + toyNumber);   
    }   
}

将输出：

Toy number in play 5  
Toy number in play after increement 6  
Toy number in main 5

在C中我可以传递toyNumber变量作为参考传递以避免阴影，即创建如下相同变量的副本：

void main(){  
    int toyNumber = 5;  
    play(toyNumber);  
    cout << "Toy number in main " << toyNumber << endl;  
}

void play(int &toyNumber){  
    cout << "Toy number in play " << toyNumber << endl;   
    toyNumber++;  
    cout << "Toy number in play after increement " << toyNumber << endl;   
}

并且C输出将是这样的：

Toy number in play 5  
Toy number in play after increement 6  
Toy number in main 6

我的问题是 - 考虑到Java is pass by value rather than pass by reference，在Java中获得与C代码相同的输出的等效代码是什么？

#1 热门回答(161 赞)

你有几个选择。最有意义的那个取决于你想要做什么。
选择1：使toyNumber成为类

class MyToy {
  public int toyNumber;
}

中的公共成员变量
然后将对MyToy的引用传递给你的方法。

void play(MyToy toy){  
    System.out.println("Toy number in play " + toy.toyNumber);   
    toy.toyNumber++;  
    System.out.println("Toy number in play after increement " + toy.toyNumber);   
}

选择2：返回值而不是通过引用传递

int play(int toyNumber){  
    System.out.println("Toy number in play " + toyNumber);   
    toyNumber++;  
    System.out.println("Toy number in play after increement " + toyNumber);   
    return toyNumber
}

这个选择需要对main中的callsite进行少量更改，以便读取，toyNumber = temp.play(toyNumber);。
选择3：使它成为类或静态变量
如果这两个函数是同一个类或类实例上的方法，则可以将toyNumber转换为类成员变量。
选择4：创建int类型的单个元素数组并传递
这被认为是hack，但有时用于从内联类调用返回值。

void play(int [] toyNumber){  
    System.out.println("Toy number in play " + toyNumber[0]);   
    toyNumber[0]++;  
    System.out.println("Toy number in play after increement " + toyNumber[0]);   
}

#2 热门回答(25 赞)

Java不是通过引用调用it is按值调用仅

但是对象类型的所有变量实际上都是指针。

因此，如果你使用Mutable对象，你将看到所需的行为

public class XYZ {

    public static void main(String[] arg) {
        StringBuilder toyNumber = new StringBuilder("5");
        play(toyNumber);
        System.out.println("Toy number in main " + toyNumber);
    }

    private static void play(StringBuilder toyNumber) {
        System.out.println("Toy number in play " + toyNumber);
        toyNumber.append(" + 1");
        System.out.println("Toy number in play after increement " + toyNumber);
    }
}

输出此代码：

run:
Toy number in play 5
Toy number in play after increement 5 + 1
Toy number in main 5 + 1
BUILD SUCCESSFUL (total time: 0 seconds)

你也可以在标准库中看到此行为。例如Collections.sort(); Collections.shuffle();这些方法不会返回新列表，但会修改它的参数对象。

List<Integer> mutableList = new ArrayList<Integer>();

    mutableList.add(1);
    mutableList.add(2);
    mutableList.add(3);
    mutableList.add(4);
    mutableList.add(5);

    System.out.println(mutableList);

    Collections.shuffle(mutableList);

    System.out.println(mutableList);

    Collections.sort(mutableList);

    System.out.println(mutableList);

输出此代码：

run:
[1, 2, 3, 4, 5]
[3, 4, 1, 5, 2]
[1, 2, 3, 4, 5]
BUILD SUCCESSFUL (total time: 0 seconds)

#3 热门回答(14 赞)

做一个

class PassMeByRef { public int theValue; }

然后将引用传递给它的实例。请注意，最好避免使用其参数改变状态的方法，尤其是在并行代码中。

原文链接