blocks|key|6024485|text|@Title本身:你不能没有“肮脏”的技巧。最简单的方法是传递一个数组或包含该字段的类。对这些的更改将会很好地反映出来。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6024486|但是您的问题完全不同:Double是不可变的，所以对它的每次更改都将返回一个新值。除了实现你自己的类之外，你对此无能为力。|6024487|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

@Title itself: You can't without "dirty" tricks. The easiest method is to pass an array - or a class containing the field. Changes to those will be reflected just fine.

But your problem is something completely different: Doubles are immutable, so every change to it will return a new value. There's nothing you can do about that, apart from implementing your own class.

blocks|key|2496594|text|包装器类型在Java中是不可变的。一旦它们被创建，它们的值就不能被改变(当然除了通过反射魔法)。你为什么要尝试做你正在做的事情呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2496595|编辑:只是为了详细说明，当您设置时：|2496596|a+=+5.0;|code-block|syntax|javascript|2496597|实际发生的事情是这样的：|2496598|a+=+new+Double(5.0);|2496599|如果double是可变的，你可以这样做|2496600|a.setValue(5.0);//+this+would+have+worked+for+mutable+objects,+but+not+in+this+case|2496601|当然，您也可以编写自己的类，但这里有很多陷阱，所以最好解释一下您的目标|2496602|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|Z|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|11|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

The wrapper types are immutable in Java. Once they are created, their value cannot be changed (except via reflection magic ofcourse). Why are you trying to do what you're doing?

Edit:
Just to elaborate, when you set:

<pre><code>a = 5.0;
</code></pre>

What actually ends up happening is something like:

<pre><code>a = new Double(5.0);
</code></pre>

If double were mutable, you could have done something like 

<pre><code>a.setValue(5.0);// this would have worked for mutable objects, but not in this case
</code></pre>

You can of course write your own class, but there are a lot of pitfalls here, so it's best to explain you're goal

blocks|key|1864850|text|你不能。在java中，您不能选择将参数传递给方法的方式(按值/按引用)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1864851|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

You can not. In java you can not select the way (by value/by ref) by which parameters will be passed to a method.

blocks|key|4608380|text|您可以尝试使用commons-lang中的MutableDouble。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4608381|但这并不是说这与按引用传递无关-您没有传递任何内容。|4608382|entityMap|0|LINK|mutability|MUTABLE|url|http://people.apache.org/~bayard/commons-lang-3.0-snapshot-api/org/apache/commons/lang/mutable/MutableDouble.html^0|L|D|L|D|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

You can try <a href="http://people.apache.org/~bayard/commons-lang-3.0-snapshot-api/org/apache/commons/lang/mutable/MutableDouble.html" rel="nofollow"><code>MutableDouble</code></a> from commons-lang.

But not that this has nothing to do with pass-by-reference - you are not passing anything.

blocks|key|6024592|text|Double在Java中是不可变的。赋值a+=+5.0并没有改变a的双精度值，它实际上是创建了一个全新的双精度对象，并改变了a持有的指针的值，以指向该新对象。如果您希望能够在不更改引用的情况下更改值，则需要创建一个可变的双精度值。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6024593|public+class+MutableDouble+{
++private+Double+value;

++public+Double+getValue()+{
++++return+value;
++}

++public+void+setValue(Double+value)+{
++++this.value+=+value;
++}

++@Override
++public+String+toString()+{
++++return+value+!=+null+?+value.toString()+:+"null";
++}
}|code-block|syntax|javascript|6024594|然后调用a.setValue(5.0)；两者都会改变。|6024595|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Double is immutable in Java. The assignment a = 5.0 is not changing the Double value of a. It is actually creating an entirely new Double object and changing the value of the pointer that a holds to point at that new object. If you want to be able to change the value without changing the reference, you need to make a mutable Double.

<pre><code>public class MutableDouble {
 private Double value;

 public Double getValue() {
 return value;
 }

 public void setValue(Double value) {
 this.value = value;
 }

 @Override
 public String toString() {
 return value != null ? value.toString() : "null";
 }
}
</code></pre>

Then you would call a.setValue(5.0); and both would change.

blocks|key|6024625|text|Java不支持指针，所以你不能直接指向a的内存(就像在C/C%2B%2B中一样)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|6024626|Java确实支持引用，但引用仅是对对象的引用。不能引用本机(内置)类型。所以当你执行时(自动装箱为你把代码转换成了下面的代码)。|6024627|双a=新双精度(3.0)；|6024628|这意味着当你执行|6024629|Double+b+=+a;|code-block|syntax|javascript|6024630|您将获得对a的对象的引用。当你选择改变一个(自动装箱最终会把你上面的代码转换成这个)|6024631|a+=+new+Double(5.0);|6024632|这不会影响b对先前创建的new+Double(3.0)的引用。换句话说，您不能通过直接操作a来影响b的引用(或者在Java中没有“远程操作”)。|offset|length|style|CODE|6024633|也就是说，还有其他的解决方案。|6024634|public+class+MutableDouble()+{

+++private+double+value;

+++public+MutableDouble(double+value)+{
+++++this.value+=+value;
+++}

+++public+double+getValue()+{
+++++return+this.value;
+++}

+++public+void+setValue(double+value)+{
+++++this.value+=+value;
+++}
+}

+MutableDouble+a+=+new+MutableDouble(3.0);
+MutableDouble+b+=+a;

+a.setValue(5.0);
+b.getValue();+//+equals+5.0|6024635|entityMap^0|0|0|0|0|0|0|0|C|F|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|14|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|15|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|16|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|17|8|@]|9|@]|A|$]]|$1|O|3|P|5|J|7|18|8|@]|9|@]|A|$K|L]]|$1|Q|3|R|5|6|7|19|8|@$S|1A|T|1B|U|V]]|9|@]|A|$]]|$1|W|3|X|5|6|7|1C|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|J|7|1D|8|@]|9|@]|A|$K|L]]|$1|10|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|11|$]]

Java doesn't support pointers, so you can't point to a's memory directly (as in C / C++).

Java does support references, but references are only references to Objects. Native (built-in) types cannot be referenced. So when you executed (autoboxing converted the code for you to the following).

Double a = new Double(3.0);

That means that when you execute

<pre><code>Double b = a;
</code></pre>

you gain a reference to a's Object. When you opt to change a (autoboxing will eventually convert your code above to this)

<pre><code>a = new Double(5.0);
</code></pre>

Which won't impact b's reference to the previously created <code>new Double(3.0)</code>. In other words, you can't impact b's reference by manipulating a directly (or there's no "action at a distance" in Java).

That said, there are other solutions

<pre><code>public class MutableDouble() {

 private double value;

 public MutableDouble(double value) {
 this.value = value;
 }

 public double getValue() {
 return this.value;
 }

 public void setValue(double value) {
 this.value = value;
 }
 }

 MutableDouble a = new MutableDouble(3.0);
 MutableDouble b = a;

 a.setValue(5.0);
 b.getValue(); // equals 5.0
</code></pre>

how can I pass a double value by reference in java?

example:

<pre><code>Double a = 3.0;
Double b = a;
System.out.println("a: "+a+" b: "+b);
a = 5.0;
System.out.println("a: "+a+" b: "+b);
</code></pre>

this code prints:

<pre><code>a: 3 b: 3
a: 5 b: 3
</code></pre>

and my problem is to get it to print:

<pre><code>a: 3 b: 3
a: 5 b: 5
</code></pre>

my goal:
I'm writing an expert system in jess, in that application a segment would have a double value for it's length, now that double value isn't in a single segment; it's in many other segments, proportionality classes..etc, all of which are referencing to it, waiting it to change so that they could possibly meet some rules.

if I can't get that double to change, I can't have a certain rule fire which contains an object that references to that double value.

java - passing a double value by reference

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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