In the following code snippet, presumably it appears that it should issue some compilation error but it doesn’t:
class Outer {
public static class Inner {
static String obj = "Inner";
}
static Optional Inner = new Optional();
//The (inner) class name and the object name are same.
}
class Optional {
String obj = "Optional";
}
public class Main {
public static void main(String[] args) {
System.out.println(Outer.Inner.obj);
//Refers to the string inside the optional class
}
}
The class Outer has a static class inside it named Inner. Additionally, it declares an object (static) of the class Optional (static Optional Inner = new Optional();)
This object and the class names (inside the class Outer) are same which is Inner. The program displays Optional. The only expression Outer.Inner.obj within main() is expected to display Inner but it doesn’t. The actual output is however Optional which is the case of the Optional class.
One way to display Inner is by changing the object name to something else.
static Optional Inner1 = new Optional();
From the output it displays, it appears that the object name (or a variable) is chosen over a type name (the class Inner) because they have the same name. What exact case is applied here?
Paragraph 6.4.2 of the Java Language Specification has some information about the rules that apply in this case.
This refers to paragraph 6.5 Determining the Meaning of a Name, which explains the rules in detail.
In your example,
Outer.Innercould refer to the type of the nested class namedInner, or the static member variableInner. The rules say the variable will be chosen over the type.