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Is Java “pass-by-reference” or “pass-by-value”?

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I always thought Java was pass-by-reference; however I've seen a couple of blog posts (for example, this blog) that claim it's not. I do not think I understand the distinction they are making.

  • Java

  • asked 3 years ago
  • B Butts

2Answer


0

Java is always pass-by-value. Unfortunately, they decided to call pointers references, thus confusing newbies. Because those references are passed by value.

It goes like this:

public static void main( String[] args ){
    Dog aDog = new Dog("Max");
    foo(aDog);

    if (aDog.getName().equals("Max")) { //true
        System.out.println( "Java passes by value." );

    } else if (aDog.getName().equals("Fifi")) {
        System.out.println( "Java passes by reference." );
    }
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true

    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 overwritten in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

Dog aDog = new Dog("Max");
foo(aDog);
aDog.getName().equals("Fifi"); // true

public void foo(Dog d) {
    d.getName().equals("Max"); // true
    d.setName("Fifi");
}
  • answered 3 years ago
  • Sunny Solu

0

I just noticed you referenced my article.

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.

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 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 someDog is set to the value 42
  • at line "AAA"
    • someDog is followed to the Dog it points to (the Dog object at address 42)
    • that Dog (the one at address 42) is asked to change his name to Max
  • at line "BBB"
    • a new Dog is created. Let's say he's at address 74
    • we assign the parameter someDog to 74
  • at line "CCC"
    • someDog is followed to the Dog it points to (the Dog object at address 74)
    • that Dog (the one at address 74) is asked to change his name to Rowlf
  • 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, you cannot change where that pointer points.

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.

  • answered 3 years ago
  • Gul Hafiz

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