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Possible Duplicate:
Type-parameterized field of a generic class becomes invisible after upgrading to Java 7
public class Test{
private String _canYouSeeMe = "yes";
<T extends Test> void genericMethod(T hey){
String s = hey._canYouSeeMe;
}
void method(Test hey){
String s = hey._canYouSeeMe;
}
}
When building against JDK 1.6 this compiles just fine but against 1.7 there is a compiler error in genericMethod(): The field Test._canYouSeeMe is not visible
The error can be resolved by making _canYouSeeMe protected rather than private, but I'm just wondering what has changed from 1.6 to 1.7
912
From the point of view of reflection, the difference between a generic type and an ordinary type is that a generic type has associated with it a set of type parameters (if it is a generic type definition) or type arguments (if it is a constructed type). A generic method differs from an ordinary method in the same way.
Many people are unsatisfied with the restrictions caused by the way generics are implemented in Java. Specifically, they are unhappy that generic type parameters are not reified: they are not available at runtime. Generics are implemented using erasure, in which generic type parameters are simply removed at runtime.
In a nutshell, generics enable types (classes and interfaces) to be parameters when defining classes, interfaces and methods. Much like the more familiar formal parameters used in method declarations, type parameters provide a way for you to re-use the same code with different inputs.
Subclasses (T) of a class (Test) never have access to the superclass' private fields. This was likely a bug in the Java 6 compiler that was fixed in Java 7.
Remember: T extends Test means that T is a subclass of Test. It does not mean that T's class is Test.class, which is the necessary condition for having private field & method access.
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