一个简单的测试,以检查一个语言是否支持通过参考,只是写一个传统的交换。

一个传统的交换方法或函数采取两个论点,并交换它们,以便转换到函数的变量在函数之外发生变化。

(非Java) 基本交换功能结构

swap(Type arg1, Type arg2) {
    Type temp = arg1;
    arg1 = arg2;
    arg2 = temp;
}

如果你能在你的语言中写出这样的方法/函数,那么

Type var1 = ...;
Type var2 = ...;
swap(var1,var2);

事实上,它交换了 var1 和 var2 变量的值,语言支持 pass-by-reference. 但 Java 不允许这样的东西,因为它支持只通过值,而不是指标或参考。

Java 以参考方式操纵对象,而所有对象变量都是参考。

例如,使用 badSwap() 方法:

    public void badSwap(int var1, int
 var2{ int temp = var1; var1 = var2; var2 =
 temp; }

public void tricky(Point arg1, Point   arg2)
{ arg1.x = 100; arg1.y = 100; Point temp = arg1; arg1 = arg2; arg2 = temp; }
public static void main(String [] args) { 

 Point pnt1 = new Point(0,0); Point pnt2
 = new Point(0,0); System.out.println("X:
 " + pnt1.x + " Y: " +pnt1.y);

     System.out.println("X: " + pnt2.x + " Y:
 " +pnt2.y); System.out.println(" ");

     tricky(pnt1,pnt2);
 System.out.println("X: " + pnt1.x + " Y:" + pnt1.y);

     System.out.println("X: " + pnt2.x + " Y: " +pnt2.y); }

如果我们执行这个主要()方法,我们会看到以下输出:

X: 0 Y: 0 X: 0 Y: 0 X: 100 Y: 100 X: 0 Y: 0

该方法成功地改变了 ofpnt1 的值,尽管它通过了值;但是,pnt1 和 pnt2 的交换失败了! 这是混乱的主要来源. 在 themain() 方法中,pnt1 和 pnt2 只是对象参考。 当你 passpnt1 和 pnt2 到 tricky() 方法时,Java 通过了值的参考,就像其他参数一样。

Java 复制并通过参考值,而不是对象. 因此,方法操纵将改变对象,因为参考指向原始对象. 但因为参考是复制,交换将失败. 如图 2 描述,方法参考交换,但不是原始参考。

public void foo(Object param)
{
  // some code in foo...
}

public void bar()
{
  Object obj = new Object();

  foo(obj);
}

它是相同的......

public void bar()
{
  Object obj = new Object();

  Object param = obj;

  // some code in foo...
}

不要考虑在这个讨论中不相关的站点。

你会遇到的最常见的运营商之一是简单的任务运营商“="......它将其右上的值归分为其左上的运营商: int cadence = 0; int speed = 0; int gear = 1; 这个运营商也可以用于对象归分对象参考。

很明显,这个运营商如何以两种不同的方式行动:分配值和分配参考;最后,当它是一个对象......第一,当它不是一个对象,即当它是一个原始的。

真相在代码中,让我们尝试一下:

public class AssignmentEvaluation
{
  static public class MyInteger
  {
    public int value = 0;
  }

  static public void main(String[] args)
  {
    System.out.println("Assignment operator evaluation using two MyInteger objects named height and width\n");

    MyInteger height = new MyInteger();
    MyInteger width  = new MyInteger();

    System.out.println("[1] Assign distinct integers to height and width values");

    height.value = 9;
    width.value  = 1;

    System.out.println("->  height is " + height.value + " and width is " + width.value + ", we are different things! \n");

    System.out.println("[2] Assign to height's value the width's value");

    height.value = width.value;

    System.out.println("->  height is " + height.value + " and width is " + width.value + ", are we the same thing now? \n");

    System.out.println("[3] Assign to height's value an integer other than width's value");

    height.value = 9;

    System.out.println("->  height is " + height.value + " and width is " + width.value + ", we are different things yet! \n");

    System.out.println("[4] Assign to height the width object");

    height = width;

    System.out.println("->  height is " + height.value + " and width is " + width.value + ", are we the same thing now? \n");

    System.out.println("[5] Assign to height's value an integer other than width's value");

    height.value = 9;

    System.out.println("->  height is " + height.value + " and width is " + width.value + ", we are the same thing now! \n");

    System.out.println("[6] Assign to height a new MyInteger and an integer other than width's value");

    height = new MyInteger();
    height.value = 1;

    System.out.println("->  height is " + height.value + " and width is " + width.value + ", we are different things again! \n");
  }
}

这就是我跑步的结果:

Assignment operator evaluation using two MyInteger objects named height and width

[1] Assign distinct integers to height and width values
->  height is 9 and width is 1, we are different things! 

[2] Assign to height's value the width's value
->  height is 1 and width is 1, are we the same thing now? 

[3] Assign to height's value an integer other than width's value
->  height is 9 and width is 1, we are different things yet! 

[4] Assign to height the width object
->  height is 1 and width is 1, are we the same thing now? 

[5] Assign to height's value an integer other than width's value
->  height is 9 and width is 9, we are the same thing now! 

[6] Assign to height a new MyInteger and an integer other than width's value
->  height is 1 and width is 9, we are different things again! 

我们不能那么确定,因为在 [6] 对象是相同的,然后我们将一个新的对象归分给其中一个,然后,对象有不同的价值! 他们现在如何区分,如果他们是相同的? 他们应该在这里也是相同的!!!

此操作员也可以用于对象分配对象参考。

短篇小说:

非原始:Java通过参考值;原始:仅值。

结局。

现在,如果你想想想什么(1)意味着,想象你有一个类的苹果:

class Apple {
    private double weight;
    public Apple(double weight) {
        this.weight = weight;
    }
    // getters and setters ...

}

然后,当你将这个类的一个例子转移到主要方法:

class Main {
    public static void main(String[] args) {
        Apple apple = new Apple(3.14);
        transmogrify(apple);
        System.out.println(apple.getWeight()+ " the goose drank wine...";

    }

    private static void transmogrify(Apple apple) {
        // does something with apple ...
        apple.setWeight(apple.getWeight()+0.55);
    }
}

哦......但你可能知道这一点,你对当你做这样的事情时会发生什么感兴趣:

class Main {
    public static void main(String[] args) {
        Apple apple = new Apple(3.14);
        transmogrify(apple);
        System.out.println("Who ate my: "+apple.getWeight()); // will it still be 3.14? 

    }

    private static void transmogrify(Apple apple) {
        // assign a new apple to the reference passed...
        apple = new Apple(2.71);
    }


}

Java 是值之通(stack memory)

它是如何工作的

首先,让我们明白,在哪里Java存储原始数据类型和对象数据类型。原始数据类型本身和对象参考存储在架子里.对象本身存储在架子里.这意味着,架子记忆存储原始数据类型以及对象的地址。

例如,使用 badSwap() 方法:

public void badSwap(int var1, int var2)
{
  int temp = var1;
  var1 = var2;
  var2 = temp;
}

当 badSwap() 返回时,作为论点过渡的变量将仍然保持其原始值。 该方法也会失败,如果我们将论点类型从 int 转换为 对象,因为 Java 通过对象参考值也。

public void tricky(Point arg1, Point arg2)
{
  arg1.x = 100;
  arg1.y = 100;
  Point temp = arg1;
  arg1 = arg2;
  arg2 = temp;
}
public static void main(String [] args)
{
  Point pnt1 = new Point(0,0);
  Point pnt2 = new Point(0,0);
  System.out.println("X: " + pnt1.x + " Y: " +pnt1.y); 
  System.out.println("X: " + pnt2.x + " Y: " +pnt2.y);
  System.out.println(" ");
  tricky(pnt1,pnt2);
  System.out.println("X: " + pnt1.x + " Y:" + pnt1.y); 
  System.out.println("X: " + pnt2.x + " Y: " +pnt2.y);  
}

如果我们执行这个主要()方法,我们会看到以下输出:

X: 0 Y: 0
X: 0 Y: 0
X: 100 Y: 100
X: 0 Y: 0

图1 转移到一个方法后,对象将至少有两种参考。

Java 复制并通过参考值,而不是对象. 因此,方法操作将改变对象,因为参考指向原始对象. 但是,因为参考是复制,交换将失败。 方法参考交换,但不是原始参考。 不幸的是,在一个方法呼叫后,你只剩下未交换的原始参考。