Initialization Sequence of this code: - When outer is created in line 1 the constructor OuterTest(String id) is called and „id“ of OuterTest is initialized to „STP“ and „Default Constructor is printed to console. - Then the instance „inner“ of the inner class is created. As inner class extends The outer class the Default-Constructor of OuterTest is called and „NonDefault Constructor“ is printed. MY QUESTION is: Why didn’t get „id“ of OuterTest the value „Default“ which is Set by the default constructor of „OuterTest“? Instead the „Outer.Test.this.id“ keeps ist old value „STP“ from the creation of the OuterTest-instance „outer“.
C:\Java\EigeneJavaProgramme>java OuterTest NonDefault-Constructor Default-Constructor InnerTest STP InnerTest Default => Question: Why is „OuterTest.this.id“ still „STP“ and not „Default“?
Joined: Jul 11, 2001
Try the following code (warning: I didn't tested it...):
This should give you false true Did that help?
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Joined: Aug 18, 2002
Let's see if I can explain what's going on. The key thing to focus on is that Inner is BOTH an inner class and a subclass. Because it's an inner class, it is automatically associated with an instance of it's outer class. Because it is a subclass, it inherits all it's superclass's properties. So when you construct the OuterTest in line 1, you get 1 instance with id set to " STP ". Now in line 2, you construct a new InnerTest via this new OuterTest, but remember that this new instance is a subclass of OuterTest, so it has it's OWN String id property, and that's what get's set to " Default ", leaving the outer instance's id unchanged. So from within the InnerTest class, OuterTest.this.id refers to the id property of the instance of Outertest that the current instance of InnerTest is implicitly associated with, and just id refers to the id property that the current instance of InnerTest inherits from the OuterTest class. Does that clarify things at all? This stuff can get very confusing.