How Can You Determine the Class of a Generic Type at Runtime?

Determining the Class of a Generic Type

Within the realm of generic programming, it can be imperative to ascertain the class of the generic type being utilized. However, as is the nature of generics, type information is typically erased at runtime, presenting a challenge in retrieving it later on.

Example:

Consider the following generic class:

public class MyGenericClass<T> {
  public void doSomething() {
    // ...
    T bean = (T)someObject.create(T.class);
    // ...
  }
}

In the example above, we attempt to create an instance of class T by passing a class literal. However, this will result in an "Illegal class literal for the type parameter T" error, as attempting to use the generic type as a class literal is forbidden.

Workaround:

To overcome this challenge, a viable workaround involves passing the generic type's class as a parameter to a static method. This method will then use the reflection API to instantiate the generic type.

public class MyGenericClass<T> {

    private final Class<T> clazz;

    public static <U> MyGenericClass<U> createMyGeneric(Class<U> clazz) {
        return new MyGenericClass<>(clazz);
    }

    protected MyGenericClass(Class<T> clazz) {
        this.clazz = clazz;
    }

    public void doSomething() {
        T instance = clazz.newInstance();
    }
}

By following this approach, we circumvent the restriction against using generic types as class literals, ensuring that the necessary type information is available at runtime. While it may be an inelegant solution, it provides a way to determine the class of a generic type in a challenging context like the one presented in the original example.

The above is the detailed content of How Can You Determine the Class of a Generic Type at Runtime?. For more information, please follow other related articles on the PHP Chinese website!