Java is Pass-by-Value, Dammit!

Can you write a traditional swap(a,b) method/function in the language?

A traditional swap method or function takes two arguments and swaps them such that variables passed into the function are changed outside the function. Its basic structure looks like

Figure 1: (Non-Java) Basic swap function structure

swap(Type arg1, Type arg2) {

    Type temp = arg1;

    arg1 = arg2;

    arg2 = temp;

}

If you can write such a method/function in your language such that calling

Figure 2: (Non-Java) Calling the swap function

Type var1 = ...;

Type var2 = ...;

swap(var1,var2);

actually switches the values of the variables var1 and var2, the language supports pass-by-reference semantics.

For example, in Pascal, you can write

Figure 3: (Pascal) Swap function

procedure swap(var arg1, arg2: SomeType);


   var

 

       temp : SomeType;

 

   begin

 

       temp := arg1;

 

       arg1 := arg2;

 

       arg2 := temp;

 

   end;

 





...


{ in some other procedure/function/program }


var

    var1, var2 : SomeType;


begin

    var1 := ...; { value "A" }

    var2 := ...; { value "B" } 

    swap(var1, var2);

    { now var1 has value "B" and var2 has value "A" }

end;

or in C++ you could write

Figure 4: (C++) Swap function

void swap(SomeType& arg1, Sometype& arg2) {

    SomeType temp = arg1;

    arg1 = arg2;

    arg2 = temp;

}


...

SomeType var1 = ...; // value "A"

SomeType var2 = ...; // value "B"

swap(var1, var2); // swaps their values!

// now var1 has value "B" and var2 has value "A"

(Please let me know if my Pascal or C++ has lapsed and I've messed up the syntax...)

But you cannot do this in Java!

---

Now the details...

The problem we're facing here is statements like

In Java, Objects are passed by reference, and primitives are passed by value.

This is half incorrect. Everyone can easily agree that primitives are passed by value; there's no such thing in Java as a pointer/reference to a primitive.

However, Objects are not passed by reference. A correct statement would be Object references are passed by value.

This may seem like splitting hairs, bit it is far from it. There is a world of difference in meaning. The following examples should help make the distinction.

In Java, take the case of

Figure 5: (Java) Pass-by-value example

public void foo(Dog d) {

    d = new Dog("Fifi"); // creating the "Fifi" dog

}






Dog aDog = new Dog("Max"); // creating the "Max" dog

// at this point, aDog points to the "Max" dog

foo(aDog);

// aDog still points to the "Max" dog

the variable passed in (aDog) is not modified! After calling foo, aDog still points to the "Max" Dog!

Many people mistakenly think/state that something like

Figure 6: (Java) Still pass-by-value...

public void foo(Dog d) { 

    d.setName("Fifi");

}

shows that Java does in fact pass objects by reference.

The mistake they make is in the definition of

Figure 7: (Java) Defining a Dog pointer

Dog d;

itself. When you write that definition, you are defining a pointer to a Dog object, not a Dog object itself.

On Pointers versus References...

The problem here is that the folks at Sun made a naming mistake.

In programming language design, a "pointer" is a variable that indirectly tracks the location of some piece of data. The value of a pointer is often the memory address of the data you're interested in. Some languages allow you to manipulate that address; others do not.

A "reference" is an alias to another variable. Any manipulation done to the reference variable directly changes the original variable.

Check out the second sentence of http://java.sun.com/docs/books/jls/third_edition/html/typesValues.html#4.3.1.

"The reference values (often just references) are pointers to these objects, and a special null reference, which refers to no object"

They emphasize "pointers" in their description... Interesting...

When they originally were creating Java, they had "pointer" in mind (you can see some remnants of this in things like

NullPointerException).

Sun wanted to push Java as a secure language, and one of Java's advantages was that it does not allow pointer arithmetic as C++ does.

They went so far as to try a different name for the concept, formally calling them "references". A big mistake and it's caused even more confusion in the process.

There's a good explanation of reference variables at http://www.cprogramming.com/tutorial/references.html. (C++ specific, but it says the right thing about the concept of a reference variable.)

The word "reference" in programming language design originally comes from how you pass data to subroutines/functions/procedures/methods. A reference parameter is an alias to a variable passed as a parameter.

In the end, Sun made a naming mistake that's caused confusion. Java has pointers, and if you accept that, it makes the way Java behaves make much more sense.

Calling Methods

Calling

Figure 8: (Java) Passing a pointer by value

foo(d);

passes the value of d to foo; it does not pass the object that d points to!

The value of the pointer being passed is similar to a memory address. Under the covers it may be a tad different, but you can think of it in exactly the same way. The value uniquely identifies some object on the heap.

However, it makes no difference how pointers are implemented under the covers. You program with them exactly the same way in Java as you would in C or C++. The syntax is just slightly different (another poor choice in Java's design; they should have used the same -> syntax for de-referencing as C++).

In Java,

Figure 9: (Java) A pointer

Dog d;

is exactly like C++'s

Figure 10: (C++) A pointer

Dog *d;

And using

Figure 11: (Java) Following a pointer and calling a method

d.setName("Fifi");

is exactly like C++'s

Figure 12: (C++) Following a pointer and calling a method

d->setName("Fifi");

To sum up: Java has pointers, and the value of the pointer is passed in. There's no way to actually pass an object itself as a parameter. You can only pass a pointer to an object.

Keep in mind, when you call

Figure 13: (Java) Even more still passing a pointer by value

foo(d);

you're not passing an object; you're passing a pointer to the object.