如何在Java中执行等效的按引用传递

  • Z时代
  • 2024-01-10
  • 分类：问答

此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);   
    }   
}

将输出以下内容：

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

在C ++中，我可以将toyNumber变量作为引用传递，以避免产生阴影，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;   
}

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

回答：

你有几种选择。最有意义的选择实际上取决于你要执行的操作。

选择1：将toyNumber设为类中的公共成员变量

class MyToy {
  public int toyNumber;
}

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
}

此选择将需要对main中的呼叫站点进行一些小的更改，使其显示为toyNumber = temp.play(toyNumber);。

选择3：将其设为类或静态变量

如果两个函数是同一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：创建一个类型为int的单个元素数组，然后传递该数组

这被认为是一种hack，但有时可用于从内联类调用中返回值。

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]);   
}

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