The terms "pass-by-value" and "pass-by-reference" have special, precisely defined meanings in computer science. These meanings differ from the intuition many people have when first hearing the terms. Much of the confusion in this discussion seems to come from this fact.
The terms "pass-by-value" and "pass-by-reference" are talking about variables. Pass-by-value means that the value of a variable is passed to a function/method. Pass-by-reference means that a reference to that variable is passed to the function. The latter gives the function a way to change the contents of the variable.
By those definitions, Java is always pass-by-value. Unfortunately, when we deal with variables holding objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.
It goes like this:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
// we pass the object to foo
foo(aDog);
// aDog variable is still pointing to the "Max" dog when foo(...) returns
aDog.getName().equals("Max"); // true
aDog.getName().equals("Fifi"); // false
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// change d inside of foo() to point to a new Dog instance construct red with name member variable set to "Fifi"
d = new Dog("Fifi");
d.getName().equals("Fifi"); // true
}
In this example, aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo by creating new Dog with name member variable set to "Fifi" because the object reference is passed by value. If the object reference was passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.
Likewise:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
foo(aDog);
// when foo(...) returns, the name of the dog has been changed to "Fifi"
aDog.getName().equals("Fifi"); // true
// but it is still the same dog:
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// this changes the name of d to be "Fifi"
d.setName("Fifi");
}
In this example, Fifi is dog’s name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.
For more information on pass by reference and pass by value, consult the following answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.
The terms "pass-by-value" and "pass-by-reference" have special, precisely defined meanings in computer science. These meanings differ from the intuition many people have when first hearing the terms. Much of the confusion in this discussion seems to come from this fact.
The terms "pass-by-value" and "pass-by-reference" are talking about variables. Pass-by-value means that the value of a variable is passed to a function/method. Pass-by-reference means that a reference to that variable is passed to the function. The latter gives the function a way to change the contents of the variable.
By those definitions, Java is always pass-by-value. Unfortunately, when we deal with variables holding objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.
It goes like this:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
// we pass the object to foo
foo(aDog);
// aDog variable is still pointing to the "Max" dog when foo(...) returns
aDog.getName().equals("Max"); // true
aDog.getName().equals("Fifi"); // false
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// change d inside of foo() to point to a new Dog instance construct red with name member variable set to "Fifi"
d = new Dog("Fifi");
d.getName().equals("Fifi"); // true
}
In this example, aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo by creating new Dog with name member variable set to "Fifi" because the object reference is passed by value. If the object reference was passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.
Likewise:
public static void main(String[] args) {
Dog aDog = new Dog("Max");
Dog oldDog = aDog;
foo(aDog);
// when foo(...) returns, the name of the dog has been changed to "Fifi"
aDog.getName().equals("Fifi"); // true
// but it is still the same dog:
aDog == oldDog; // true
}
public static void foo(Dog d) {
d.getName().equals("Max"); // true
// this changes the name of d to be "Fifi"
d.setName("Fifi");
}
In this example, Fifi is dog’s name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.
For more information on pass by reference and pass by value, consult the following answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.
I'm the author of the blog post you're talking about. To clarify a few things:
The Java Spec says that everything in Java is pass-by-value. There is no such thing as "pass-by-reference" in Java.
The key to understanding this is that something like
Dog myDog;
is not a Dog; it's actually a pointer to a Dog. The use of the term "reference" in Java is very misleading and is what causes most of the confusion here. What they call "references" act/feel more like what we'd call "pointers" in most other languages.
What that means, is when you have
Dog myDog = new Dog("Rover");
foo(myDog);
you're essentially passing the address of the created Dog object to the foo method.
(I say essentially because Java pointers/references aren't direct addresses, but it's easiest to think of them that way.)
Suppose the Dog object resides at memory address 42. This means we pass 42 to the method.
if the Method were defined as
public void foo(Dog someDog) {
someDog.setName("Max"); // AAA
someDog = new Dog("Fifi"); // BBB
someDog.setName("Rowlf"); // CCC
}
let's look at what's happening.
- the parameter
someDogis set to the value 42 - at line "AAA"
someDogis followed to theDogit points to (theDogobject at address 42)- that
Dog(the one at address 42) is asked to change his name to Max
- at line "BBB"
- a new
Dogis created. Let's say he's at address 74 - we assign the parameter
someDogto 74
- a new
- at line "CCC"
- someDog is followed to the
Dogit points to (theDogobject at address 74) - that
Dog(the one at address 74) is asked to change his name to Rowlf
- someDog is followed to the
- then, we return
Now let's think about what happens outside the method:
Did myDog change?
There's the key.
Keeping in mind that myDog is a pointer, and not an actual Dog, the answer is NO. myDog still has the value 42; it's still pointing to the original Dog (but note that because of line "AAA", its name is now "Max" - still the same Dog; myDog's value has not changed.)
It's perfectly valid to follow an address and change what's at the end of it; that does not change the variable, however.
Java works exactly like C. You can assign a pointer, pass the pointer to a method, follow the pointer in the method and change the data that was pointed to. However, the caller will not see any changes you make to where that pointer points. (In a language with pass-by-reference semantics, the method function can change the pointer and the caller will see that change.)
In C++, Ada, Pascal and other languages that support pass-by-reference, you can actually change the variable that was passed.
If Java had pass-by-reference semantics, the foo method we defined above would have changed where myDog was pointing when it assigned someDog on line BBB.
Think of reference parameters as being aliases for the variable passed in. When that alias is assigned, so is the variable that was passed in.
A discussion in the comments warrants some clarification...
In C, you can write
void swap(int *x, int *y) {
int t = *x;
*x = *y;
*y = t;
}
int x = 1;
int y = 2;
swap(&x, &y);
This is not a special case in C. Both languages use pass-by-value semantics. Here the call site is creating additional data structure to assist the function to access and manipulate data.
The function is being passed pointers to data, and follows those pointers to access and modify that data.
A similar approach in Java, where the caller sets up assisting structure, might be:
void swap(int[] x, int[] y) {
int temp = x[0];
x[0] = y[0];
y[0] = temp;
}
int[] x = {1};
int[] y = {2};
swap(x, y);
(or if you wanted both examples to demonstrate features the other language doesn't have, create a mutable IntWrapper class to use in place of the arrays)
In these cases, both C and Java are simulating pass-by-reference. They're still both passing values (pointers to ints or arrays), and following those pointers inside the called function to manipulate the data.
Pass-by-reference is all about the function declaration/definition, and how it handles its parameters. Reference semantics apply to every call to that function, and the call site only needs to pass variables, no additional data structure.
These simulations require the call site and the function to cooperate. No doubt it's useful, but it's still pass-by-value.
Can someone explain pass by value in Java please?
Is Java pass-by-value or pass-by-reference?
How Kotlin pass variables, by reference or by value?
Are arrays pass by reference or pass by value?
In java except the primitive data types like int, char everything else is pass by reference is a wrong concept. Everything is pass by value. In case of objects you are passing the reference of the object to the new object . The references are passed by value. For ex arr[ ] = new int[5]; Here arr is a variable pointing to a memory location of array of int. Now consider,
Fn(int[ ] arr2){ .... }
Call fn(arr);
Here the reference of arr that is the memory location of where the array of integer starts will be given to arr2.
Hope this clears your confusion.
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I've read multiple explanations online but I'm still having trouble understanding. I understand that Java is pass by value but from what I understand objects are passed by reference. My programming teacher even had trouble explaining this. Can someone please give me a clear explanation of this, in layman's terms, as if I had never programmed before?
Also, if someone could give some of the pros and cons of pass by value and pass by reference that would be great too, and what languages are pass by value and which are pass by reference?