Java: how do I get a class literal from a generic type?
Asked Answered
C

9

265

Typically, I've seen people use the class literal like this:

Class<Foo> cls = Foo.class;

But what if the type is generic, e.g. List? This works fine, but has a warning since List should be parameterized:

Class<List> cls = List.class

So why not add a <?>? Well, this causes a type mismatch error:

Class<List<?>> cls = List.class

I figured something like this would work, but this is just a plain ol' syntax error:

Class<List<Foo>> cls = List<Foo>.class

How can I get a Class<List<Foo>> statically, e.g. using the class literal?

I could use @SuppressWarnings("unchecked") to get rid of the warnings caused by the non-parameterized use of List in the first example, Class<List> cls = List.class, but I'd rather not.

Any suggestions?

Carve answered 5/3, 2010 at 23:33 Comment(0)
A
195

You can't due to type erasure.

Java generics are little more than syntactic sugar for Object casts. To demonstrate:

List<Integer> list1 = new ArrayList<Integer>();
List<String> list2 = (List<String>)list1;
list2.add("foo"); // perfectly legal

The only instance where generic type information is retained at runtime is with Field.getGenericType() if interrogating a class's members via reflection.

All of this is why Object.getClass() has this signature:

public final native Class<?> getClass();

The important part being Class<?>.

To put it another way, from the Java Generics FAQ:

Why is there no class literal for concrete parameterized types?

Because parameterized type has no exact runtime type representation.

A class literal denotes a Class object that represents a given type. For instance, the class literal String.class denotes the Class object that represents the type String and is identical to the Class object that is returned when method getClass is invoked on a String object. A class literal can be used for runtime type checks and for reflection.

Parameterized types lose their type arguments when they are translated to byte code during compilation in a process called type erasure . As a side effect of type erasure, all instantiations of a generic type share the same runtime representation, namely that of the corresponding raw type . In other words, parameterized types do not have type representation of their own. Consequently, there is no point in forming class literals such as List<String>.class , List<Long>.class and List<?>.class , since no such Class objects exist. Only the raw type List has a Class object that represents its runtime type. It is referred to as List.class.

Addlepated answered 5/3, 2010 at 23:39 Comment(8)
Aha! This makes more sense. I figured that the class literal on a generic wouldn't even make much sense, but I had no idea that it was due to this. Thanks!Carve
List<Integer> list1 = new ArrayList<Integer>(); List<String> list2 = (List<String>)list1; list2.add("foo"); // perfectly legal You can;t do that in Java, you get type mismatch compilation error !Lazybones
so... what do I do if I need one?Nail
You can always fool the compiler by List<String> list2 = (List<String>) (Object) list1;Jeconiah
Yet another "It just works in C#, but not in Java" for me. I am deserialising a JSON object, and typeof(List<MyClass>) works perfectly fine in C#, but List<MyClass>.class is a syntax error in Java. Yes, there is a logical explanation for that as usual as Cletus wrote, but I always wonder why all those things just work in C#.Keelung
@ChristopherFrancisco see the comment by kftse that follows yours, and more directly, the separate Answer by slaurent.Determine
what do you mean it perfectly legal? That part of the code doesn't compile?Touching
@DamnVegetables If you're just trying to deserialize you can deserialize using MyClass[].class as that actually exists and it will deserialize properly.Homology
P
83

There are no Class literals for parameterized types, however there are Type objects that correctly define these types.

See java.lang.reflect.ParameterizedType - http://java.sun.com/j2se/1.5.0/docs/api/java/lang/reflect/ParameterizedType.html

Google's Gson library defines a TypeToken class that allows to simply generate parameterized types and uses it to spec json objects with complex parameterized types in a generic friendly way. In your example you would use:

Type typeOfListOfFoo = new TypeToken<List<Foo>>(){}.getType()

I intended to post links to the TypeToken and Gson classes javadoc but Stack Overflow won't let me post more than one link since I'm a new user, you can easily find them using Google search

Piloting answered 8/3, 2010 at 4:34 Comment(1)
With this I was able create a class with a generic E and then use clzz = new TypeToken<E>(){}.getRawType(); to later iterate over similarly structured enums with clzz.getEnumConstants() and then finally use refection to call member methods a la Method method = clzz.getDeclaredMethod("getSomeFoo"); so much win! Thank YOU!Cunningham
B
78

You can manage it with a double cast :

@SuppressWarnings("unchecked") Class<List<Foo>> cls = (Class<List<Foo>>)(Object)List.class

Bede answered 10/6, 2015 at 11:29 Comment(6)
By changing the second cast from Object to Class you can probably save the overhead of a (pointless) checked runtime cast.Minaminabe
@Minaminabe Using Class instead of Object, as you suggest, seems more meaningful but it doesn't remove the need of the @SuppressWarnings("unchecked") annotation, it even add a new warning : Class is a raw type. References to generic type Class<T> should be parameterizedGord
You can use Class<?>: (Class<List<Foo>>)(Class<?>)List.classFaulk
@Faulk I see you are correct when I try that... What are the arguments for using either (Object) or (Class<?>)?Determine
This answer is totally pointless. The reason OP wanted to parameterize the class path was because he got a unchecked warning. This answer doesn't change / improve any of that. OP even states in his question that he doesn't want to use SuppressWarnings...Minne
@Minne That might be true, but it solved the problem I had - a constructor of my supertype needed a Class<List<T>> where I have defined the exact type of T.Bailiwick
T
9

To expound on cletus' answer, at runtime all record of the generic types is removed. Generics are processed only in the compiler and are used to provide additional type safety. They are really just shorthand that allows the compiler to insert typecasts at the appropriate places. For example, previously you'd have to do the following:

List x = new ArrayList();
x.add(new SomeClass());
Iterator i = x.iterator();
SomeClass z = (SomeClass) i.next();

becomes

List<SomeClass> x = new ArrayList<SomeClass>();
x.add(new SomeClass());
Iterator<SomeClass> i = x.iterator();
SomeClass z = i.next();

This allows the compiler to check your code at compile-time, but at runtime it still looks like the first example.

Tireless answered 5/3, 2010 at 23:44 Comment(3)
Thanks for the additional explanation--my understanding of generics is so much clearer now that I realize they're not a runtime mechanism. :)Carve
In my opinion, this just means that generic were implemented in a mediocre manner by Sun, hopefully Oracle fixes this some day. C#'s implementation of generic is much much much better (Anders is godlike)Witte
@MarcelValdezOrozco AFAIK, in Java they implemented it that way because they wanted old code (pre-1.5) to work on new JVMs without any issues. Seems it's a very smart design decision which cares about compatibility. I don't think there's anything mediocre in that.Inappropriate
V
5

You could use a helper method to get rid of @SuppressWarnings("unchecked") all over a class.

@SuppressWarnings("unchecked")
private static <T> Class<T> generify(Class<?> cls) {
    return (Class<T>)cls;
}

Then you could write

Class<List<Foo>> cls = generify(List.class);

Other usage examples are

  Class<Map<String, Integer>> cls;

  cls = generify(Map.class);

  cls = TheClass.<Map<String, Integer>>generify(Map.class);

  funWithTypeParam(generify(Map.class));

public void funWithTypeParam(Class<Map<String, Integer>> cls) {
}

However, since it is rarely really useful, and the usage of the method defeats the compiler's type checking, I would not recommend to implement it in a place where it is publicly accessible.

Vishinsky answered 9/11, 2016 at 23:23 Comment(1)
This is quite useful for implementing a concrete subclass of com.fasterxml.jackson.databind.deser.std.StdDeserializer with a generic type such as a collection.Bailiwick
G
5

The Java Generics FAQ and therefore also cletus' answer sound like there is no point in having Class<List<T>>, however the real problem is that this is extremely dangerous:

@SuppressWarnings("unchecked")
Class<List<String>> stringListClass = (Class<List<String>>) (Class<?>) List.class;

List<Integer> intList = new ArrayList<>();
intList.add(1);
List<String> stringList = stringListClass.cast(intList);
// Surprise!
String firstElement = stringList.get(0);

The cast() makes it look as if it is safe, but in reality it is not safe at all.


Though I don't get where there can't be List<?>.class = Class<List<?>> since this would be pretty helpful when you have a method which determines the type based on the generic type of a Class argument.

For getClass() there is JDK-6184881 requesting to switch to using wildcards, however it does not look like this change will be performed (very soon) since it is not compatible with previous code (see this comment).

Gismo answered 4/1, 2019 at 0:36 Comment(0)
S
2

Well as we all know that it gets erased. But it can be known under some circumstances where the type is explicitly mentioned in the class hierarchy:

import java.lang.reflect.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.stream.Collectors;

public abstract class CaptureType<T> {
    /**
     * {@link java.lang.reflect.Type} object of the corresponding generic type. This method is useful to obtain every kind of information (including annotations) of the generic type.
     *
     * @return Type object. null if type could not be obtained (This happens in case of generic type whose information cant be obtained using Reflection). Please refer documentation of {@link com.types.CaptureType}
     */
    public Type getTypeParam() {
        Class<?> bottom = getClass();
        Map<TypeVariable<?>, Type> reifyMap = new LinkedHashMap<>();

        for (; ; ) {
            Type genericSuper = bottom.getGenericSuperclass();
            if (!(genericSuper instanceof Class)) {
                ParameterizedType generic = (ParameterizedType) genericSuper;
                Class<?> actualClaz = (Class<?>) generic.getRawType();
                TypeVariable<? extends Class<?>>[] typeParameters = actualClaz.getTypeParameters();
                Type[] reified = generic.getActualTypeArguments();
                assert (typeParameters.length != 0);
                for (int i = 0; i < typeParameters.length; i++) {
                    reifyMap.put(typeParameters[i], reified[i]);
                }
            }

            if (bottom.getSuperclass().equals(CaptureType.class)) {
                bottom = bottom.getSuperclass();
                break;
            }
            bottom = bottom.getSuperclass();
        }

        TypeVariable<?> var = bottom.getTypeParameters()[0];
        while (true) {
            Type type = reifyMap.get(var);
            if (type instanceof TypeVariable) {
                var = (TypeVariable<?>) type;
            } else {
                return type;
            }
        }
    }

    /**
     * Returns the raw type of the generic type.
     * <p>For example in case of {@code CaptureType<String>}, it would return {@code Class<String>}</p>
     * For more comprehensive examples, go through javadocs of {@link com.types.CaptureType}
     *
     * @return Class object
     * @throws java.lang.RuntimeException If the type information cant be obtained. Refer documentation of {@link com.types.CaptureType}
     * @see com.types.CaptureType
     */
    public Class<T> getRawType() {
        Type typeParam = getTypeParam();
        if (typeParam != null)
            return getClass(typeParam);
        else throw new RuntimeException("Could not obtain type information");
    }


    /**
     * Gets the {@link java.lang.Class} object of the argument type.
     * <p>If the type is an {@link java.lang.reflect.ParameterizedType}, then it returns its {@link java.lang.reflect.ParameterizedType#getRawType()}</p>
     *
     * @param type The type
     * @param <A>  type of class object expected
     * @return The Class<A> object of the type
     * @throws java.lang.RuntimeException If the type is a {@link java.lang.reflect.TypeVariable}. In such cases, it is impossible to obtain the Class object
     */
    public static <A> Class<A> getClass(Type type) {
        if (type instanceof GenericArrayType) {
            Type componentType = ((GenericArrayType) type).getGenericComponentType();
            Class<?> componentClass = getClass(componentType);
            if (componentClass != null) {
                return (Class<A>) Array.newInstance(componentClass, 0).getClass();
            } else throw new UnsupportedOperationException("Unknown class: " + type.getClass());
        } else if (type instanceof Class) {
            Class claz = (Class) type;
            return claz;
        } else if (type instanceof ParameterizedType) {
            return getClass(((ParameterizedType) type).getRawType());
        } else if (type instanceof TypeVariable) {
            throw new RuntimeException("The type signature is erased. The type class cant be known by using reflection");
        } else throw new UnsupportedOperationException("Unknown class: " + type.getClass());
    }

    /**
     * This method is the preferred method of usage in case of complex generic types.
     * <p>It returns {@link com.types.TypeADT} object which contains nested information of the type parameters</p>
     *
     * @return TypeADT object
     * @throws java.lang.RuntimeException If the type information cant be obtained. Refer documentation of {@link com.types.CaptureType}
     */
    public TypeADT getParamADT() {
        return recursiveADT(getTypeParam());
    }

    private TypeADT recursiveADT(Type type) {
        if (type instanceof Class) {
            return new TypeADT((Class<?>) type, null);
        } else if (type instanceof ParameterizedType) {
            ArrayList<TypeADT> generic = new ArrayList<>();
            ParameterizedType type1 = (ParameterizedType) type;
            return new TypeADT((Class<?>) type1.getRawType(),
                    Arrays.stream(type1.getActualTypeArguments()).map(x -> recursiveADT(x)).collect(Collectors.toList()));
        } else throw new UnsupportedOperationException();
    }

}

public class TypeADT {
    private final Class<?> reify;
    private final List<TypeADT> parametrized;

    TypeADT(Class<?> reify, List<TypeADT> parametrized) {
        this.reify = reify;
        this.parametrized = parametrized;
    }

    public Class<?> getRawType() {
        return reify;
    }

    public List<TypeADT> getParameters() {
        return parametrized;
    }
}

And now you can do things like:

static void test1() {
        CaptureType<String> t1 = new CaptureType<String>() {
        };
        equals(t1.getRawType(), String.class);
    }

    static void test2() {
        CaptureType<List<String>> t1 = new CaptureType<List<String>>() {
        };
        equals(t1.getRawType(), List.class);
        equals(t1.getParamADT().getParameters().get(0).getRawType(), String.class);
    }


    private static void test3() {
            CaptureType<List<List<String>>> t1 = new CaptureType<List<List<String>>>() {
            };
            equals(t1.getParamADT().getRawType(), List.class);
        equals(t1.getParamADT().getParameters().get(0).getRawType(), List.class);
    }

    static class Test4 extends CaptureType<List<String>> {
    }

    static void test4() {
        Test4 test4 = new Test4();
        equals(test4.getParamADT().getRawType(), List.class);
    }

    static class PreTest5<S> extends CaptureType<Integer> {
    }

    static class Test5 extends PreTest5<Integer> {
    }

    static void test5() {
        Test5 test5 = new Test5();
        equals(test5.getTypeParam(), Integer.class);
    }

    static class PreTest6<S> extends CaptureType<S> {
    }

    static class Test6 extends PreTest6<Integer> {
    }

    static void test6() {
        Test6 test6 = new Test6();
        equals(test6.getTypeParam(), Integer.class);
    }



    class X<T> extends CaptureType<T> {
    }

    class Y<A, B> extends X<B> {
    }

    class Z<Q> extends Y<Q, Map<Integer, List<List<List<Integer>>>>> {
    }

    void test7(){
        Z<String> z = new Z<>();
        TypeADT param = z.getParamADT();
        equals(param.getRawType(), Map.class);
        List<TypeADT> parameters = param.getParameters();
        equals(parameters.get(0).getRawType(), Integer.class);
        equals(parameters.get(1).getRawType(), List.class);
        equals(parameters.get(1).getParameters().get(0).getRawType(), List.class);
        equals(parameters.get(1).getParameters().get(0).getParameters().get(0).getRawType(), List.class);
        equals(parameters.get(1).getParameters().get(0).getParameters().get(0).getParameters().get(0).getRawType(), Integer.class);
    }




    static void test8() throws IllegalAccessException, InstantiationException {
        CaptureType<int[]> type = new CaptureType<int[]>() {
        };
        equals(type.getRawType(), int[].class);
    }

    static void test9(){
        CaptureType<String[]> type = new CaptureType<String[]>() {
        };
        equals(type.getRawType(), String[].class);
    }

    static class SomeClass<T> extends CaptureType<T>{}
    static void test10(){
        SomeClass<String> claz = new SomeClass<>();
        try{
            claz.getRawType();
            throw new RuntimeException("Shouldnt come here");
        }catch (RuntimeException ex){

        }
    }

    static void equals(Object a, Object b) {
        if (!a.equals(b)) {
            throw new RuntimeException("Test failed. " + a + " != " + b);
        }
    }

More info here. But again, it is almost impossible to retrieve for:

class SomeClass<T> extends CaptureType<T>{}
SomeClass<String> claz = new SomeClass<>();

where it gets erased.

Serviette answered 27/4, 2015 at 10:0 Comment(1)
This is also the workaround used by JAX-RS, cf. GenericEntity and GenericType.Syncrisis
A
1

Due to the exposed fact that Class literals doesn't have generic type information, I think you should assume that it will be impossible to get rid of all the warnings. In a way, using Class<Something> is the same as using a collection without specifying the generic type. The best I could come out with was:

private <C extends A<C>> List<C> getList(Class<C> cls) {
    List<C> res = new ArrayList<C>();
    // "snip"... some stuff happening in here, using cls
    return res;
}

public <C extends A<C>> List<A<C>> getList() {
    return getList(A.class);
}
Arnitaarno answered 8/3, 2010 at 14:34 Comment(0)
H
0

What's not possible in Java is still possible if you add some Kotlin to the mix.

Create a Kotlin file with the following content:

fun <T> getGenericListType(): Class<List<T>> {
    return emptyList<T>().javaClass
}

fun <T> getGenericListSuperType(): Class<MutableList<in T>> {
    return mutableListOf<T>().javaClass
}

fun <T> getGenericListExtendsType(): Class<MutableList<out T>> {
    return mutableListOf<T>().javaClass
}

Calling these methods from Java will give you, respectively:

  1. a Class<List<T>>,
  2. a Class<List<? super T>>, and
  3. a Class<List<? extends T>>.
Higherup answered 17/5, 2023 at 10:1 Comment(0)

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