Does Java Use Pass-By-Value Semantics?


If Java passes objects by passing the reference variable instead, does that mean Java uses pass-by-reference for objects? Not exactly, although you’ll often hear and read that it does. Java is actually pass-by-value for all variables running within a single VM. Pass-by-value means pass-by-variable-value. And that means, pass-by-copy-of-the-variable! (There’s that word copy again!)

It makes no difference if you’re passing primitive or reference variables, you are always passing a copy of the bits in the variable. So for a primitive variable, you’re passing a copy of the bits representing the value. For example, if you pass an int variable with the value of 3, you’re passing a copy of the bits representing 3. The called method then gets its own copy of the value, to do with it what it likes.

And if you’re passing an object reference variable, you’re passing a copy of the bits representing the reference to an object. The called method then gets its own copy of the reference variable, to do with it what it likes. But because two identical reference variables refer to the exact same object, if the called method modifies the object (by invoking setter methods, for example), the caller will see that the object the caller’s original variable refers to has also been changed.

The bottom line on pass-by-value: the called method can’t change the caller’s variable, although for object reference variables, the called method can change the object the variable referred to. What’s the difference between changing the variable and changing the object? For object references, it means the called method can’t reassign the caller’s original reference variable and make it refer to a different object, or null. For example, in the following code fragment,
                          void bar() {
                                Foo f = new Foo();
                                doStuff(f);
                           }
                           void doStuff(Foo g) {
                                  g.setName(“Boo”);
                                 g = new Foo();
                            }
reassigning g does not reassign f! At the end of the bar() method, two Foo objects have been created, one referenced by the local variable f and one referenced by the local (argument) variable g. Because the doStuff() method has a copy of the reference variable, it has a way to get to the original Foo object, for instance to call the setName() method. But, the doStuff() method does not have a way to get to the f reference variable. So doStuff() can change values within the object f refers to, but doStuff() can’t change the actual contents (bit pattern) of f. In other words, doStuff() can change the state of the object that f refers to, but it can’t make f refer to a different object! 

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