Java Spec说,Java中的所有东西都是按值传递的。 Java中没有“传递参考”这样的东西。


Dog myDog;

不是狗; 它实际上是一个指向狗的指针


Dog myDog = new Dog("Rover");





public void foo(Dog someDog) {
    someDog.setName("Max");     // AAA
    someDog = new Dog("Fifi");  // BBB
    someDog.setName("Rowlf");   // CCC


  • 参数someDog被设置为值42
  • 在“AAA”行
    • someDog Dog跟随它指向的Dog (地址42处的Dog对象)
    • Dog (地址42处的Dog )被要求将他的名字改为Max
  • 在“BBB”行
    • 一个新的Dog被创建。 假设他在地址74
    • 我们将参数someDog分配给74
  • 在“CCC”行
    • 有些Dog跟随它指向的Dog (地址为74的Dog对象)
    • Dog (地址为74的那个)被要求将他的名字改为Rowlf
  • 那么,我们回来




记住myDog是一个指针 ,而不是真正的Dog ,答案是NO。 myDog仍然有值42; 它仍然指向原来的Dog (但请注意,由于“AAA”行,它的名字现在是“最大” - 仍然是狗; myDog的值没有改变。)

遵循地址并改变结尾是完全有效的; 但不会改变变量。

Java的工作方式与C完全一样。您可以指定一个指针,将指针传递给方法,跟随方法中的指针并更改指向的数据。 但是,您无法更改指针指向的位置。

在C ++,Ada,Pascal和其他支持通过引用的语言中,您实际上可以更改传递的变量。




我一直认为Java是通过引用的 ,但是我见过一些博客文章(例如, 这个博客 ),声称它不是。 我不认为我理解他们的区别。


You can never pass by reference in Java, and one of the ways that is obvious is when you want to return more than one value from a method call. Consider the following bit of code in C++:

void getValues(int& arg1, int& arg2) {
    arg1 = 1;
    arg2 = 2;
void caller() {
    int x;
    int y;
    getValues(x, y);
    cout << "Result: " << x << " " << y << endl;

Sometimes you want to use the same pattern in Java, but you can't; at least not directly. Instead you could do something like this:

void getValues(int[] arg1, int[] arg2) {
    arg1[0] = 1;
    arg2[0] = 2;
void caller() {
    int[] x = new int[1];
    int[] y = new int[1];
    getValues(x, y);
    System.out.println("Result: " + x[0] + " " + y[0]);

As was explained in previous answers, in Java you're passing a pointer to the array as a value into getValues . That is enough, because the method then modifies the array element, and by convention you're expecting element 0 to contain the return value. Obviously you can do this in other ways, such as structuring your code so this isn't necessary, or constructing a class that can contain the return value or allow it to be set. But the simple pattern available to you in C++ above is not available in Java.


在C ++中: 注意:错误代码 - 内存泄漏! 但它证明了这一点。

void cppMethod(int val, int &ref, Dog obj, Dog &objRef, Dog *objPtr, Dog *&objPtrRef)
    val = 7; // Modifies the copy
    ref = 7; // Modifies the original variable
    obj.SetName("obj"); // Modifies the copy of Dog passed
    objRef.SetName("objRef"); // Modifies the original Dog passed
    objPtr->SetName("objPtr"); // Modifies the original Dog pointed to 
                               // by the copy of the pointer passed.
    objPtr = new Dog("newObjPtr");  // Modifies the copy of the pointer, 
                                   // leaving the original object alone.
    objPtrRef->SetName("objRefPtr"); // Modifies the original Dog pointed to 
                                    // by the original pointer passed. 
    objPtrRef = new Dog("newObjPtrRef"); // Modifies the original pointer passed

int main()
    int a = 0;
    int b = 0;
    Dog d0 = Dog("d0");
    Dog d1 = Dog("d1");
    Dog *d2 = new Dog("d2");
    Dog *d3 = new Dog("d3");
    cppMethod(a, b, d0, d1, d2, d3);
    // a is still set to 0
    // b is now set to 7
    // d0 still have name "d0"
    // d1 now has name "objRef"
    // d2 now has name "objPtr"
    // d3 now has name "newObjPtrRef"


public static void javaMethod(int val, Dog objPtr)
   val = 7; // Modifies the copy
   objPtr.SetName("objPtr") // Modifies the original Dog pointed to 
                            // by the copy of the pointer passed.
   objPtr = new Dog("newObjPtr");  // Modifies the copy of the pointer, 
                                  // leaving the original object alone.

public static void main()
    int a = 0;
    Dog d0 = new Dog("d0");
    javaMethod(a, d0);
    // a is still set to 0
    // d0 now has name "objPtr"


In Java only references are passed and are passed by value:

Java arguments are all passed by value (the reference is copied when used by the method) :

In the case of primitive types, Java behaviour is simple: The value is copied in another instance of the primitive type.

In case of Objects, this is the same: Object variables are pointers (buckets) holding only Object's address that was created using the "new" keyword, and are copied like primitive types.

The behaviour can appear different from primitive types: Because the copied object-variable contains the same address (to the same Object) Object's content/members might still be modified within a method and later access outside, giving the illusion that the (containing) Object itself was passed by reference.

"String" Objects appear to be a perfect counter-example to the urban legend saying that "Objects are passed by reference":

In effect, within a method you will never be able, to update the value of a String passed as argument:

A String Object, holds characters by an array declared final that can't be modified. Only the address of the Object might be replaced by another using "new". Using "new" to update the variable, will not let the Object be accessed from outside, since the variable was initially passed by value and copied.

我不敢相信没有人提到芭芭拉·利斯科夫。 当她在1974年设计CLU时,她遇到了同样的术语问题,她发明了通过共享 (也称为通过对象共享按对象呼叫呼叫的术语,一个参考“。



The crux of the matter is that the word reference in the expression "pass by reference" means something completely different from the usual meaning of the word reference in Java.

Usually in Java reference means aa reference to an object . But the technical terms pass by reference/value from programming language theory is talking about a reference to the memory cell holding the variable , which is something completely different.

Java copies the reference by value. So if you change it to something else (eg, using new ) the reference does not change outside the method. For native types, it is always pass by value.

I always think of it as "pass by copy". It is a copy of the value be it primitive or reference. If it is a primitive it is a copy of the bits that are the value and if it is an Object it is a copy of the reference.

public class PassByCopy{
    public static void changeName(Dog d){ = "Fido";
    public static void main(String[] args){
        Dog d = new Dog("Maxx");
        System.out.println("name= "+;
        System.out.println("name= "+;
class Dog{
    public String name;
    public Dog(String s){ = s;

output of java PassByCopy:

name= Maxx
name= Fido

Primitive wrapper classes and Strings are immutable so any example using those types will not work the same as other types/objects.

The distinction, or perhaps just the way I remember as I used to be under the same impression as the original poster is this: Java is always pass by value. All objects( in Java, anything except for primitives) in Java are references. These references are passed by value.

Java is always pass by value, not pass by reference

First of all, we need to understand what pass by value and pass by reference are.

Pass by value means that you are making a copy in memory of the actual parameter's value that is passed in. This is a copy of the contents of the actual parameter .

Pass by reference (also called pass by address) means that a copy of the address of the actual parameter is stored .

Sometimes Java can give the illusion of pass by reference. Let's see how it works by using the example below:

public class PassByValue {
    public static void main(String[] args) {
        Test t = new Test(); = "initialvalue";
        new PassByValue().changeValue(t);

    public void changeValue(Test f) { = "changevalue";

class Test {
    String name;

The output of this program is:


Let's understand step by step:

Test t = new Test();

As we all know it will create an object in the heap and return the reference value back to t. For example, suppose the value of t is 0x100234 (we don't know the actual JVM internal value, this is just an example) .

new PassByValue().changeValue(t);

When passing reference t to the function it will not directly pass the actual reference value of object test, but it will create a copy of t and then pass it to the function. Since it is passing by value , it passes a copy of the variable rather than the actual reference of it. Since we said the value of t was 0x100234 , both t and f will have the same value and hence they will point to the same object.

If you change anything in the function using reference f it will modify the existing contents of the object. That is why we got the output changevalue , which is updated in the function.

To understand this more clearly, consider the following example:

public class PassByValue {
    public static void main(String[] args) {
        Test t = new Test(); = "initialvalue";
        new PassByValue().changeRefence(t);

    public void changeRefence(Test f) {
        f = null;

class Test {
    String name;

Will this throw a NullPointerException ? No, because it only passes a copy of the reference. In the case of passing by reference, it could have thrown a NullPointerException , as seen below:

Hopefully this will help.


第一步请从头脑中清除以'p'开头的词,尤其是如果你来自其他编程语言。 Java和'p'不能写在同一本书,论坛甚至txt中。


  • 学生 :师父,这是否意味着Java是通过引用?
  • 主人 :蚱蜢,号


  1. 一个变量保存指示JVM如何到达内存中引用的对象(Heap)的位。
  2. 将参数传递给方法时, 您不会传递引用变量,而是引用变量中的位的副本 。 就像这样:3bad086a。 3bad086a代表一种获取传递对象的方法。
  3. 所以你只是通过3bad086a,它是参考的价值。
  4. 你传递的是引用的值,而不是引用本身(而不是对象)。
  5. 该值实际上是COPIED并赋予该方法


1. Person person;
2. person = new Person("Tom");
3. changeName(person);
5. //I didn't use Person person below as an argument to be nice
6. static void changeName(Person anotherReferenceToTheSamePersonObject) {
7.     anotherReferenceToTheSamePersonObject.setName("Jerry");
8. }


  • 变量人员在#1行中创建,并在开始时为空。
  • 一个新的Person对象在第2行中创建,存储在内存中,变量Person被赋予对Person对象的引用。 就是它的地址。 假设3bad086a。
  • 持有对象地址的变量人员被传递给第3行中的函数。
  • 在#4线你可以听到沉默的声音
  • 检查第5行的评论
  • 一个方法局部变量 - anotherReferenceToTheSamePersonObject - 被创建,然后在#6行出现魔法:
    • 变量/引用被逐位复制并传递给函数内的另一个参考对象。
    • 没有创建Person的新实例。
    • person ”和“ anotherReferenceToTheSamePersonObject ”都保持3bad086a的相同值。
    • 不要尝试这个,但人== anotherReferenceToTheSamePersonObject将是真实的。
    • 这两个变量都有引用的IDENTICAL COPIES,它们都引用同一个Person对象,堆上的SAME对象而不是COPY。



如果你没有得到它,那么请相信我并记住,最好说Java是通过价值传递的 。 那么, 通过参考值 。 哦,更好的是传递可变值的副本! ;)



  • 如果它是基本数据类型,这些位将包含基本数据类型本身的值。
  • 如果它是一个Object,这些位将包含告诉JVM如何到达Object的地址的值。

Java是通过值传递的,因为在一个方法中,您可以尽可能多地修改引用的对象,但不管您尝试多么努力,都将永远无法修改将继续引用的传递变量(而不是p _ _ _ _ _ _ _)相同的对象无论如何!

上面的changeName函数将永远不能修改传入引用的实际内容(位值)。 换句话说,changeName不能让Person人引用另一个Object。


A reference is always a value when represented, no matter what language you use.

Getting an outside of the box view, let's look at Assembly or some low level memory management. At the CPU level a reference to anything immediately becomes a value if it gets written to memory or to one of the CPU registers. (That is why pointer is a good definition. It is a value, which has a purpose at the same time).

Data in memory has a Location and at that location there is a value (byte,word, whatever). In Assembly we have a convenient solution to give a Name to certain Location (aka variable), but when compiling the code, the assembler simply replaces Name with the designated location just like your browser replaces domain names with IP addresses.

Down to the core it is technically impossible to pass a reference to anything in any language without representing it (when it immediately becomes a value).

Lets say we have a variable Foo, its Location is at the 47th byte in memory and its Value is 5. We have another variable Ref2Foo which is at 223rd byte in memory, and its value will be 47. This Ref2Foo might be a technical variable, not explicitly created by the program. If you just look at 5 and 47 without any other information, you will see just two Values . If you use them as references then to reach to 5 we have to travel:

(Name)[Location] -> [Value at the Location]
(Ref2Foo)[223]  -> 47
(Foo)[47]       -> 5

This is how jump-tables work.

If we want to call a method/function/procedure with Foo's value, there are a few possible way to pass the variable to the method, depending on the language and its several method invocation modes:

  1. 5 gets copied to one of the CPU registers (ie. EAX).
  2. 5 gets PUSHd to the stack.
  3. 47 gets copied to one of the CPU registers
  4. 47 PUSHd to the stack.
  5. 223 gets copied to one of the CPU registers.
  6. 223 gets PUSHd to the stack.

In every cases above a value - a copy of an existing value - has been created, it is now upto the receiving method to handle it. When you write "Foo" inside the method, it is either read out from EAX, or automatically dereferenced , or double dereferenced, the process depends on how the language works and/or what the type of Foo dictates. This is hidden from the developer until she circumvents the dereferencing process. So a reference is a value when represented, because a reference is a value that has to be processed (at language level).

Now we have passed Foo to the method:

  • in case 1. and 2. if you change Foo ( Foo = 9 ) it only affects local scope as you have a copy of the Value. From inside the method we cannot even determine where in memory the original Foo was located.
  • in case 3. and 4. if you use default language constructs and change Foo ( Foo = 11 ), it could change Foo globally (depends on the language, ie. Java or like Pascal's procedure findMin(x, y, z: integer; var m : integer); ). However if the language allows you to circumvent the dereference process, you can change 47 , say to 49 . At that point Foo seems to have been changed if you read it, because you have changed the local pointer to it. And if you were to modify this Foo inside the method ( Foo = 12 ) you will probably FUBAR the execution of the program (aka. segfault) because you will write to a different memory than expected, you can even modify an area that is destined to hold executable program and writing to it will modify running code (Foo is now not at 47 ). BUT Foo's value of 47 did not change globally, only the one inside the method, because 47 was also a copy to the method.
  • in case 5. and 6. if you modify 223 inside the method it creates the same mayhem as in 3. or 4. (a pointer, pointing to a now bad value, that is again used as a pointer) but this is still a local problem, as 223 was copied . However if you are able to dereference Ref2Foo (that is 223 ), reach to and modify the pointed value 47 , say, to 49 , it will affect Foo globally , because in this case the methods got a copy of 223 but the referenced 47 exists only once, and changing that to 49 will lead every Ref2Foo double-dereferencing to a wrong value.

Nitpicking on insignificant details, even languages that do pass-by-reference will pass values to functions, but those functions know that they have to use it for dereferencing purposes. This pass-the-reference-as-value is just hidden from the programmer because it is practically useless and the terminology is only pass-by-reference .

Strict pass-by-value is also useless, it would mean that a 100 Mbyte array should have to be copied every time we call a method with the array as argument, therefore Java cannot be stricly pass-by-value. Every language would pass a reference to this huge array (as a value) and either employs copy-on-write mechanism if that array can be changed locally inside the method or allows the method (as Java does) to modify the array globally (from the caller's view) and a few languages allows to modify the Value of the reference itself.

So in short and in Java's own terminology, Java is pass-by-value where value can be: either a real value or a value that is a representation of a reference .

To make a long story short, Java objects have some very peculiar properties.

In general, Java has primitive types ( int , bool , char , double , etc) that are passed directly by value. Then Java has objects (everything that derives from java.lang.Object ). Objects are actually always handled through a reference (a reference being a pointer that you can't touch). That means that in effect, objects are passed by reference, as the references are normally not interesting. It does however mean that you cannot change which object is pointed to as the reference itself is passed by value.

Does this sound strange and confusing? Let's consider how C implements pass by reference and pass by value. In C, the default convention is pass by value. void foo(int x) passes an int by value. void foo(int *x) is a function that does not want an int a , but a pointer to an int: foo(&a) . One would use this with the & operator to pass a variable address.

Take this to C++, and we have references. References are basically (in this context) syntactic sugar that hide the pointer part of the equation: void foo(int &x) is called by foo(a) , where the compiler itself knows that it is a reference and the address of the non-reference a should be passed. In Java, all variables referring to objects are actually of reference type, in effect forcing call by reference for most intends and purposes without the fine grained control (and complexity) afforded by, for example, C++.

As far as I know, Java only knows call by value. This means for primitive datatypes you will work with an copy and for objects you will work with an copy of the reference to the objects. However I think there are some pitfalls; for example, this will not work:

public static void swap(StringBuffer s1, StringBuffer s2) {
    StringBuffer temp = s1;
    s1 = s2;
    s2 = temp;

public static void main(String[] args) {
    StringBuffer s1 = new StringBuffer("Hello");
    StringBuffer s2 = new StringBuffer("World");
    swap(s1, s2);

This will populate Hello World and not World Hello because in the swap function you use copys which have no impact on the references in the main. But if your objects are not immutable you can change it for example:

public static void appendWorld(StringBuffer s1) {
    s1.append(" World");

public static void main(String[] args) {
    StringBuffer s = new StringBuffer("Hello");

This will populate Hello World on the command line. If you change StringBuffer into String it will produce just Hello because String is immutable. 例如:

public static void appendWorld(String s){
    s = s+" World";

public static void main(String[] args) {
    String s = new String("Hello");

However you could make a wrapper for String like this which would make it able to use it with Strings:

class StringWrapper {
    public String value;

    public StringWrapper(String value) {
        this.value = value;

public static void appendWorld(StringWrapper s){
    s.value = s.value +" World";

public static void main(String[] args) {
    StringWrapper s = new StringWrapper("Hello");

edit: i believe this is also the reason to use StringBuffer when it comes to "adding" two Strings because you can modifie the original object which u can't with immutable objects like String is.