How does wrapper class object initialisation works?

Question

In Java

Integer i=10;

Integer as wrapper class. How it is initialised while it's an object reference only? Could anybody explain that?

Answer 1

The statement

Integer i = 10;

is short for:

Integer i; // Variable declaration
i = 10; // Variable assignment

So first of all Java creates a variable called i which is allowed to refer to instances of type Integer.

After that the right side is executed, 10 is an integer literal of type int (see JLS§3.10.1). Then Java tries to assign the int to your Integer variable i. The types differ, for a regular case the assignment would not be possible.

However, Java is able to automatically convert int to Integer and vice versa. This is called autoboxing, Java can do this for all primitives and their wrapper (see JLS§5.1.7).

So Java converts your int 10 to the wrapper type Integer by using Integer.valueOf(10). In fact, since the number is small (-128 to +127) the method returns a cached Integer out of an internal pool instead of creating a new instance.

So after Java magic your code is:

Integer i = Integer.valueOf(10);

which is a valid assignment as the right side is also an Integer and not an int anymore.

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Actually the usage of Integer.valueOf is not described in the JLS. It only says that Java must be able to automatically convert int to Integer (from JLS§5.1.7):

If p is a value of type int, then boxing conversion converts p into a reference r of class and type Integer, such that r.intValue() == p.

But most JVM implementations use that method (you can see that in the resulting byte code).