Java è "pass-by-reference" o "pass-by-value"?



Answers

Ho appena notato che hai fatto riferimento al mio articolo .

La specifica Java dice che tutto in Java è pass-by-value. Non esiste una cosa come "pass-by-reference" in Java.

La chiave per capire questo è qualcosa di simile

Dog myDog;

non è un cane; in realtà è un puntatore a un cane.

Cosa significa, è quando hai

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

stai essenzialmente passando l' indirizzo dell'oggetto Dog creato al metodo foo .

(Dico essenzialmente perché i puntatori Java non sono indirizzi diretti, ma è più facile pensarli in quel modo)

Supponiamo che l'oggetto Dog risieda all'indirizzo di memoria 42. Ciò significa che passiamo 42 al metodo.

se il metodo è stato definito come

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

diamo un'occhiata a cosa sta succedendo.

  • il parametro someDog è impostato sul valore 42
  • alla riga "AAA"
    • someDog è seguito al Dog cui punta (l'oggetto Dog all'indirizzo 42)
    • a quel Dog (quello all'indirizzo 42) viene chiesto di cambiare il suo nome in Max
  • alla riga "BBB"
    • viene creato un nuovo Dog Diciamo che è all'indirizzo 74
    • assegniamo il parametro someDog a 74
  • alla linea "CCC"
    • someDog è seguito al Dog cui punta (l'oggetto Dog all'indirizzo 74)
    • quel Dog (quello all'indirizzo 74) è invitato a cambiare il suo nome in Rowlf
  • poi, torniamo

Ora pensiamo a cosa succede al di fuori del metodo:

myDog cambiato?

C'è la chiave.

Tenendo presente che myDog è un puntatore e non un vero Dog , la risposta è NO. myDog ha ancora il valore 42; sta ancora puntando al Dog originale (ma nota che a causa della linea "AAA", il suo nome ora è "Max" - è ancora lo stesso Cane; il valore di myDog non è cambiato.)

È perfettamente valido seguire un indirizzo e cambiare ciò che è alla fine di esso; questo non cambia la variabile, comunque.

Java funziona esattamente come C. È possibile assegnare un puntatore, passare il puntatore a un metodo, seguire il puntatore nel metodo e modificare i dati puntati. Tuttavia, non puoi cambiare dove punta quel puntatore.

In C ++, Ada, Pascal e in altre lingue che supportano il pass-by-reference, puoi effettivamente cambiare la variabile che è stata passata.

Se Java avesse una semantica pass-by-reference, il metodo foo che abbiamo definito sopra sarebbe cambiato dove myDog stava puntando quando ha assegnato un someDog on line BBB.

Pensa ai parametri di riferimento come alias per la variabile passata. Quando viene assegnato quell'alias, lo è anche la variabile passata.

Question

Ho sempre pensato che Java fosse un riferimento pass-by , tuttavia ho visto un paio di post del blog (ad esempio, questo blog ) che affermano che non lo è. Non credo di capire la distinzione che stanno facendo.

Qual è la spiegazione?




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. Per esempio:

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.




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.




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.




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.




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.




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;
}

The output of this program is:

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:

Hopefully this will help.




Solo per mostrare il contrasto, confronta i seguenti frammenti C++ e Java :

In C ++: Nota: codice errato - perdite di memoria! Ma dimostra il 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"
}

In 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 ha solo i due tipi di passaggio: per valore per i tipi predefiniti e per valore del puntatore per i tipi di oggetto.




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.




Java passa i riferimenti in base al valore.

Quindi non puoi cambiare il riferimento che viene passato.




Non posso credere che nessuno abbia menzionato Barbara Liskov ancora. Quando ha progettato CLU nel 1974, si è imbattuta in questo stesso problema terminologico e ha inventato il termine chiamata per condivisione (noto anche come chiamata per condivisione di oggetti e chiamata per oggetto ) per questo caso specifico di "call by value dove il valore è a reference".




Questo ti darà alcuni spunti su come Java funzioni davvero al punto che nella tua prossima discussione su Java che passa per riferimento o che passi per valore sorriderà :-)

Passo uno per favore cancella dalla tua mente quella parola che inizia con 'p' "_ _ _ _ _ _" ", specialmente se provieni da altri linguaggi di programmazione. Java e 'p' non possono essere scritti nello stesso libro, forum o anche txt.

La seconda fase ricorda che quando si passa un oggetto in un metodo si passa il riferimento all'oggetto e non l'oggetto stesso.

  • Studente : Master, significa che Java è pass-by-reference?
  • Maestro : cavalletta, n.

Ora pensa a che cosa fa / è un riferimento / variabile di un oggetto:

  1. Una variabile contiene i bit che indicano alla JVM come raggiungere l'oggetto di riferimento in memoria (Heap).
  2. Quando si passano argomenti a un metodo NON si sta passando la variabile di riferimento, ma una copia dei bit nella variabile di riferimento . Qualcosa del genere: 3bad086a. 3bad086a rappresenta un modo per arrivare all'oggetto passato.
  3. Quindi stai solo passando a 3bad086a che è il valore del riferimento.
  4. Stai passando il valore del riferimento e non il riferimento stesso (e non l'oggetto).
  5. Questo valore è in realtà COPIATO e dato al metodo .

Di seguito (per favore non provare a compilare / eseguire questo ...):

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. }

Che succede?

  • La persona variabile viene creata nella riga n. 1 ed è nulla all'inizio.
  • Un nuovo Oggetto Persona viene creato nella riga n. 2, memorizzato nella memoria e alla persona variabile viene assegnato il riferimento all'oggetto Persona. Cioè, il suo indirizzo. Diciamo 3bad086a.
  • La persona variabile che detiene l'indirizzo dell'oggetto viene passata alla funzione nella riga 3.
  • Alla riga 4 puoi ascoltare il suono del silenzio
  • Controlla il commento sulla linea # 5
  • Una variabile locale del metodo - anotherReferenceToTheSamePersonObject - viene creata e poi viene la magia nella riga n. 6:
    • La variabile / persona di riferimento viene copiata bit per bit e passata a anotherReferenceToTheSamePersonObject all'interno della funzione.
    • Non vengono create nuove istanze di Person.
    • Sia " persona " che " anotherReferenceToTheSamePersonObject " hanno lo stesso valore di 3bad086a.
    • Non provare questo, ma person == anotherReferenceToTheSamePersonObject sarebbe vero.
    • Entrambe le variabili hanno COPIE IDENTICHE del riferimento e si riferiscono entrambi allo stesso oggetto Person, allo SAME Object sull'Heap e NON A COPY.

Un'immagine vale più di mille parole:

Nota che le frecce anotherReferenceToTheSamePersonObject sono dirette verso l'Oggetto e non verso la persona variabile!

Se non l'hai capito, fidati di me e ricorda che è meglio dire che Java è un valore . Bene, passa per valore di riferimento . Oh bene, ancora meglio è il valore della copia pass-by-the-variable! ;)

Ora sentitevi liberi di odiarmi, ma osservate che, dato ciò, non vi è alcuna differenza tra il passaggio di tipi di dati primitivi e Oggetti quando si parla di argomenti del metodo.

Si passa sempre una copia dei bit del valore del riferimento!

  • Se si tratta di un tipo di dati primitivo, questi bit conterranno il valore del tipo di dati primitivi stesso.
  • Se è un oggetto, i bit conterranno il valore dell'indirizzo che dice alla JVM come raggiungere l'oggetto.

Java è pass-by-value perché all'interno di un metodo puoi modificare l'oggetto di riferimento quanto vuoi ma non importa quanto duramente ci provi non sarai mai in grado di modificare la variabile passata che manterrà il riferimento (non p _ _ _ _ _ _ _) lo stesso oggetto, non importa cosa!

La funzione changeName precedente non sarà mai in grado di modificare il contenuto effettivo (i valori di bit) del riferimento passato. In altre parole changeName non può rendere Person persona riferirsi a un altro oggetto.

Ovviamente puoi tagliarlo brevemente e dire semplicemente che Java è un valore pass-by!




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.




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++.




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