¿Java es "pass-by-reference" o "pass-by-value"?


Answers

Me acabo de dar cuenta de que has hecho referencia a mi artículo .

Java Spec dice que todo en Java es pass-by-value. No existe tal cosa como "pass-by-reference" en Java.

La clave para entender esto es que algo así como

Dog myDog;

no es un perro; en realidad es un puntero a un perro.

Lo que eso significa, es cuando tienes

Dog myDog = new Dog("Rover");
foo(myDog);

esencialmente está pasando la dirección del objeto Dog creado al método foo .

(Digo esencialmente porque los punteros de Java no son direcciones directas, pero es más fácil pensar en ellos de esa manera)

Supongamos que el objeto Dog reside en la dirección de memoria 42. Esto significa que pasamos 42 al método.

si el Método se definió como

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

echemos un vistazo a lo que está sucediendo.

  • el parámetro someDog se establece en el valor 42
  • en la línea "AAA"
    • someDog se sigue al Dog que apunta (el objeto Dog en la dirección 42)
    • se le pide a ese Dog (el de la dirección 42) que cambie su nombre a Max
  • en la línea "BBB"
    • Se crea un nuevo Dog . Digamos que está en la dirección 74
    • asignamos el parámetro someDog a 74
  • en la línea "CCC"
    • SomeDog se sigue al Dog que apunta (el objeto Dog en la dirección 74)
    • a ese Dog (el de la dirección 74) se le pide que cambie su nombre a Rowlf
  • entonces, volvemos

Ahora pensemos en lo que ocurre fuera del método:

¿ myDog ?

Ahí está la clave.

Teniendo en cuenta que myDog es un puntero , y no un Dog real, la respuesta es NO. myDog todavía tiene el valor 42; todavía apunta al Dog original (pero tenga en cuenta que debido a la línea "AAA", su nombre ahora es "Máx." - sigue siendo el mismo perro; el valor de myDog no ha cambiado).

Es perfectamente válido seguir una dirección y cambiar lo que está al final; eso no cambia la variable, sin embargo.

Java funciona exactamente como C. Puede asignar un puntero, pasar el puntero a un método, seguir el puntero en el método y cambiar los datos que se señalaron. Sin embargo, no puede cambiar dónde apunta ese puntero.

En C ++, Ada, Pascal y otros lenguajes que soportan pass-by-reference, en realidad puede cambiar la variable que se pasó.

Si Java tuviera una semántica de paso a paso, el método foo que definimos anteriormente habría cambiado a donde myDog cuando asignó someDog en la línea BBB.

Piense en los parámetros de referencia como alias para la variable pasada. Cuando se asigna ese alias, también lo es la variable que se pasó.

Question

Siempre pensé que Java pasaba de referencia , sin embargo, he visto un par de publicaciones en el blog (por ejemplo, este blog ) que afirman que no lo es. No creo entender la distinción que están haciendo.

¿Cuál es la explicación?




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 .




No puedo creer que nadie haya mencionado a Barbara Liskov todavía. Cuando diseñó CLU en 1974, se encontró con este mismo problema de terminología e inventó el término llamada mediante el uso compartido (también conocido como llamada por intercambio de objetos y llamada por objeto ) para este caso específico de "llamada por valor donde el valor es una referencia".




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




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.




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.




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++.




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.




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.




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);
    System.out.println(s1);
    System.out.println(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");
    appendWorld(s);
    System.out.println(s);
}

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

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

public static void main(String[] args) {
    String s = new String("Hello");
    appendWorld(s);
    System.out.println(s);
}

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");
    appendWorld(s);
    System.out.println(s.value);
}

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.




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();
        t.name = "initialvalue";
        new PassByValue().changeValue(t);
        System.out.println(t.name);
    }

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

class Test {
    String name;
}

El resultado de este programa es:

changevalue

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();
        t.name = "initialvalue";
        new PassByValue().changeRefence(t);
        System.out.println(t.name);
    }

    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:

Espero que esto ayude.




Esto te dará una idea de cómo funciona realmente Java hasta el punto de que en tu próxima discusión sobre Java pasando por referencia o pasando por valor, solo sonreirás :-)

Paso uno borre de su mente esa palabra que comienza con 'p' "_ _ _ _ _ _ _", especialmente si proviene de otros lenguajes de programación. Java y 'p' no se pueden escribir en el mismo libro, foro o incluso txt.

El segundo paso recuerda que cuando pasas un Objeto a un método estás pasando la referencia del Objeto y no el Objeto mismo.

  • Estudiante : Maestro, ¿esto significa que Java es pasado por referencia?
  • Maestro : Saltamontes, No.

Ahora piense en lo que hace / es una referencia / variable de un objeto:

  1. Una variable contiene los bits que le dicen a la JVM cómo llegar al Objeto referenciado en la memoria (Heap).
  2. Al pasar argumentos a un método NO pasa la variable de referencia, sino una copia de los bits en la variable de referencia . Algo como esto: 3bad086a. 3bad086a representa una forma de llegar al objeto pasado.
  3. Así que solo está pasando 3bad086a que es el valor de la referencia.
  4. Está pasando el valor de la referencia y no la referencia en sí (y no el objeto).
  5. Este valor es realmente COPIADO y dado al método .

A continuación, (no intente compilar / ejecutar esto ...):

1. Person person;
2. person = new Person("Tom");
3. changeName(person);
4.
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. }

¿Lo que pasa?

  • La persona variable se crea en la línea n. ° 1 y es nula al principio.
  • Se crea un nuevo objeto de persona en la línea 2, almacenado en la memoria, y la persona variable recibe la referencia al objeto de persona. Es decir, su dirección. Digamos 3bad086a.
  • La persona variable que tiene la dirección del objeto se pasa a la función en la línea # 3.
  • En la línea # 4 puedes escuchar el sonido del silencio
  • Verifique el comentario en la línea # 5
  • Se crea una variable local de método - anotherReferenceToTheSamePersonObject - y luego viene la magia en la línea # 6:
    • La persona de referencia / variable se copia bit a bit y se pasa a otro ObjectReferenceToTheSamePersonObject dentro de la función.
    • No se crean nuevas instancias de Persona.
    • Tanto " persona " como "otro objeto ReferenceReferenceToTheSamePerson " tienen el mismo valor que 3bad086a.
    • No intente esto, pero la persona == anotherReferenceToTheSamePersonObject sería verdadera.
    • Ambas variables tienen COPIAS IDENTICAS de la referencia y ambas se refieren al mismo objeto de persona, el MISMO objeto en el montón y NO A UNA COPIA.

Una imagen vale mas que mil palabras:

Tenga en cuenta que las flechas anotherReferenceToTheSamePersonObject se dirigen hacia el objeto y no hacia la persona variable.

Si no lo obtuviste, solo confía en mí y recuerda que es mejor decir que Java pasa de valor . Bueno, pase por valor de referencia . ¡Oh, bueno, aún mejor es pasar-por-copiar-el-valor-variable! ;)

Ahora siéntete libre de odiarme, pero ten en cuenta que, dado esto, no hay diferencia entre pasar tipos de datos primitivos y Objetos cuando se habla de argumentos de métodos.

¡Siempre pasa una copia de los bits del valor de la referencia!

  • Si se trata de un tipo de datos primitivo, estos bits contendrán el valor del tipo de datos primitivo en sí.
  • Si es un Objeto, los bits contendrán el valor de la dirección que le dice a la JVM cómo llegar al Objeto.

Java es paso por valor porque dentro de un método puede modificar el objeto al que hace referencia tanto como lo desee, pero no importa cuánto lo intente, nunca podrá modificar la variable pasada que seguirá haciendo referencia (no p _ _ _ _ _ _ _) el mismo Objeto sin importar qué!

La función changeName anterior nunca podrá modificar el contenido real (los valores de bit) de la referencia aprobada. En otras palabras changeName no puede hacer que Persona persona se refiera a otro Object.

Por supuesto, puede resumirlo y solo decir que Java es pasable de valor.




Java pasa referencias por valor.

Entonces no puedes cambiar la referencia que se pasa.




Para mostrar el contraste, compare los siguientes fragmentos de C++ y Java :

En C ++: Nota: código incorrecto - ¡pérdidas de memoria! Pero demuestra el punto.

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"
}

En Java,

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"
}

Java solo tiene los dos tipos de aprobación: por valor para los tipos incorporados y por valor del puntero para los tipos de objeto.




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