Final answer:
Java generics use type erasure, enable bounded type parameters, and do not support primitive types directly, unlike C++ templates, which maintain type information, offer template specialization, and work with primitive types.
Step-by-step explanation:
The generic parameters in a Java 5.0 generic method differ substantially from those in C++ in several fundamental ways. First, Java generics are implemented using type erasure, which means that the generic type information is removed during compilation. This is different from C++ templates which are a compile-time construct and maintain type information. As a result, Java generics introduce certain constraints, such as not being able to instantiate generic types or perform certain types of type checks at runtime.
Furthermore, Java generics add bounded type parameters, allowing developers to restrict the kinds of types that can be used as arguments in a generic type. For example, using <T extends Comparable<T>> to ensure that the type passed to the generic class implements Comparable. This is similar to C++'s template specialization but is often considered to be more limited in scope.
Lastly, Java's generics do not support primitive types directly; they require the use of wrapper classes, while C++ templates can work with primitives directly. This affects performance and usability in high-performance computing scenarios where primitives are preferred.