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Inner class constructor

 
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The above code prints
STP
STP
In the above code inner class constructor InnerTest() doesn't get invoked? Shouldn't it print
STP
Default
Thanks
veena
 
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Hi Veena,
I think the inner class constructor does get called. I added the following line in the Inner class constructor and it was printed.
System.out.println("inner class constructor");

Thanks,
Cathy.
 
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In the InnerTest instance there is a id field inherited, however Outer.this.id is also accesible. The latter is a field of the containing class, it will be initialized by a constructor of its class: new OuterTest("STP"). The former would be intialized by InnerTest constructor, either directly (id="something" or indirectly by calling super();
Try System.out.println(OuterTest.this.id +"\t"+ id);
and you will see
STP Default
STP Default
 
Cathy Song
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Hi Jose,
I cannot figure out how calling super() changes the inherited id variable? Can you please explain this concept.
Thanks,
Cathy.
 
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super() calls the no-args (default) constructor of the superclass, in this case, OuterTest().
Inside this OuterTest() constructor, the variable id of InnerTest is assigned to "Default", because 'this' is referring to the current object.
So OuterTest's id is still "STP".
Hope you understood.
Ana
 
Cathy Song
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But the output is the same even if I take the "this" out..
OuterTest ( ) {
//this.id = " Default " ;
id = " Default " ;
}
Now the id variable is refering to Inner or outer? I am totally lost..
 
Jose Botella
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Welcome to the Ranch Ana.
____________________________________________________
Cathy, within a constructor or instance method id is equivalent to this.id .
this is the hidden argument to instance methods representing the object on which the method was called: "aCar.go();" within go() this refers to aCar.
this in a constructor is the object being initialized, that is the object created by new.
 
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