Isn't there any way to find the class-type of a generic?
if (T instanceof String) {
// do something...
}
The above definitely does not compile.
4 Answers
Generics are a compile time feature. Generics add checks at compile time which may not have any meaning at runtime. This is one example. You can only check the type of the object referenced which could be a super type in code. If you want to pass the type T you have do this explicitly.
void someMethod(Class<T> tClass) {
if(String.class.isAssignableFrom(tClass))
or
void someMethod(Class<T> tClass, T tArg) {
Note: the type might not be the same,
someMethod(Number.class, 1);
It won't compile because T is not a variable, but a place holder for a class that is defined at runtime. Here's a quick sample:
public class Test<T> {
public void something(T arg) {
if (arg instanceof String) {
System.out.println("Woot!");
}
}
public static void main(String[] args) {
Test<String> t = new Test<String>();
t.something("Hello");
}
}
if you have subclass
public class SomeClass extends SomeSubclass<String>{}
and
public class SomeSubclass<T> {}
then there is a way to discover type of T by executing code
Type t = getClass().getGenericSuperclass()
if (t instanceof ParameterizedType) {
Type[] actualTypeArguments = ((ParameterizedType)t).getActualTypeArguments()
// in simple cases actualTypeArguments will contain Classes, since Class implements Type
}
if your case are a bit more complex (? extends String)` take a look at org.ormunit.entity.AEntityAccessor#extractClass
If you have specific field you can just check it like below:
private <T> String someMethod(T genericElement)
{
if (String.class.isInstance(genericElement))
{
return (String) genericElement;
}
...
