问题
I have question related to Java inner classes.
Is there a way to access top level class A from top level class Main that define inner class A?
Below is sample code demonstrating the problem:
class A { // Outer Class A
{
System.out.println("A outer");
}
}
class B { // Outer Class B
{
System.out.println("B outer");
}
}
public class Main {
class A { // Inner Class A
{
System.out.println("A inner");
}
}
public void newA() {
class A { // Local Class A
{
System.out.println("A local");
}
}
new A();
}
public static void main(String[] args) {
new Main().newA(); // prints "A local"
new Main().new A(); // prints "A inner"
//new A(); // compiler error: No enclosing instance of type Main is Accessible.
new B(); // but this works and prints "B outer"
}
}
回答1:
Use the fully qualified class name:
new com.mypackage.A();
回答2:
If your classes are in packages, I'd expect that to be the simplest way of achieving this. I don't believe there's any equivalent of the C# global::
for Java, to force this sort of thing.
Ultimately though, I'd just try to change the class names to avoid the problem happening in the first place. That's usually a better approach than using workarounds.
回答3:
You should provide different package names for your two A
s so that they are clearly different types, then you can reference them by their full names.
For example:
my.package.for.toplevel.A outerA = new my.package.for.toplevel.A(); // prints "A outer"
// if this is the
// package
Or even better, use two different names for two distinct classes.
回答4:
The name of the inner class is said to shadow the name of the top-level class. The top-level class cannot be referenced by its simple name in the scope of the inner class; the top-level class can only be referenced via a qualified name.
If the top-level class is in the default package (in which case its canonical name is the same as its simple name), you can still access it via reflection.
回答5:
Yep, just use packages or avoid naming them the same from the beginning. I wonder what's your reason to name them the same anyway?
To explain your error, it comes from the fact that it tries to access the inner A, but since that class is not declared as static and there's no Main instance available, it can't create a non-static inner A that requires a reference to parent Main instance.
回答6:
Assumed the classes are in the default package. To resolve naming conflict in this case needs either rename the conflicting classes or use the named package. Then use FQCN to resolve the issue.
来源:https://stackoverflow.com/questions/5897853/java-inner-class-with-the-same-name-as-other-top-level-class