If you create a generic class in Java (the class has generic type parameters), can you use generic methods (the method takes generic type parameters)?
Consider the following example:
public class MyClass { public <K> K doSomething(K k){ return k; } } public class MyGenericClass<T> { public <K> K doSomething(K k){ return k; } public <K> List<K> makeSingletonList(K k){ return Collections.singletonList(k); } }
As you would expect with a generic method, I can call doSomething(K)
on instances of MyClass
with any object:
MyClass clazz = new MyClass(); String string = clazz.doSomething("String"); Integer integer = clazz.doSomething(1);
However, if I try to use instances of MyGenericClass
without specifying a generic type, I calling doSomething(K)
returns an Object
, regardless of what K
was passed in:
MyGenericClass untyped = new MyGenericClass(); // this doesn't compile - "Incompatible types. Required: String, Found: Object" String string = untyped.doSomething("String");
Oddly, it will compile if the return type is a generic class - e.g. List<K>
(Actually, this can be explained - see answer below):
MyGenericClass untyped = new MyGenericClass(); List<String> list = untyped.makeSingletonList("String"); // this compiles
Also, it will compile if the generic class is typed, even if only with wildcards:
MyGenericClass<?> wildcard = new MyGenericClass(); String string = wildcard.doSomething("String"); // this compiles
Is there a good reason why calling a generic method in an untyped generic class shouldn't work?
Is there some clever trick relating to generic classes and generic methods that I am missing?
EDIT:
To clarify, I would expect an untyped or raw-typed generic class not to honour the generic class's type parameters (because they haven't been provided). However, it's not clear to my why an untyped or raw-typed generic class would mean that generic methods are not honoured.
It transpires that this issue has already been raised on SO, c.f. this question. The answers to this explain that when a class is untyped / in its raw-form, all generics are removed from the class - including typing of generic methods.
However, there isn't really an explanation as to why this is the case. So allow me to clarify my question:
EDIT - discussion of JLS:
It has been suggested (in answer to the previous SO question and to this question) that this is treated in JLS 4.8, which states:
The type of a constructor (§8.8), instance method (§8.4, §9.4), or non-static field (§8.3) M of a raw type C that is not inherited from its superclasses or superinterfaces is the raw type that corresponds to the erasure of its type in the generic declaration corresponding to C.
It is clear to me how this relates to an untyped class - the class generic types are replaced with the erasure types. If the class generics are bound, then the erasure type corresponds to those bounds. If the they are not bound, then the erasure type is Object - e.g.
// unbound class types public class MyGenericClass<T> { public T doSomething(T t) { return t; } } MyGenericClass untyped = new MyGenericClass(); Object t = untyped.doSomething("String"); // bound class types public class MyBoundedGenericClass<T extends Number> { public T doSomething(T t) { return t; } } MyBoundedGenericClass bounded = new MyBoundedGenericClass(); Object t1 = bounded.doSomething("String"); // does not compile Number t2 = bounded.doSomething(1); // does compile
Whilst generic methods are instance methods, it is not clear to me that JLS 4.8 applies to generic methods. The generic method's type (<K>
in earlier example) is not untyped, as it's type is determined by the method parameters - only the class is untyped / raw-typed.
Generic methods are methods that introduce their own type parameters. This is similar to declaring a generic type, but the type parameter's scope is limited to the method where it is declared. Static and non-static generic methods are allowed, as well as generic class constructors.
Generics Work Only with Reference Types: When we declare an instance of a generic type, the type argument passed to the type parameter must be a reference type. We cannot use primitive data types like int, char. Test<int> obj = new Test<int>(20);
Generic methods allow type parameters to be used to express dependencies among the types of one or more arguments to a method and/or its return type. If there isn't such a dependency, a generic method should not be used. It is possible to use both generic methods and wildcards in tandem. Here is the method Collections.
'for backwards compatibility' seems a sufficient reason for the type erasure of class generic types - it is needed e.g. to allow you to return an untyped List and pass it to some legacy code. The extension of this to generic methods seems like a tricky sub-case.
The JLS snippet from 4.8 (which you quote) covers constructors, instance methods and member fields - generic methods are just a particular case of instance methods in general. So it seems your case is covered by this snippet.
Adapting JLS 4.8 to this specific case :
The type of a generic method is the raw type that corresponds to the erasure of its type in the generic declaration corresponding to C.
(here the 'type' of the method would include all parameter and return types). If you interpret 'erasure' as 'erasing all generics', then this does seem to match the observed behaviour, although it is not very intuitive or even useful. It almost seems like an overzealous consistency, to erase all generics, rather than just generic class parameters (although who am I to second guess the designers).
Perhaps there could be problems where the class generic parameters interact with the method generic parameters - in your code they are fully independent, but you could imagine other cases where they are assigned / mixed together. I think it's worth pointing out that use of raw types are not recommended, as per the JLS :
The use of raw types is allowed only as a concession to compatibility of legacy code. The use of raw types in code written after the introduction of genericity into the Java programming language is strongly discouraged. It is possible that future versions of the Java programming language will disallow the use of raw types
Some of the thinking of the java developers is apparent here :
http://bugs.sun.com/view_bug.do?bug_id=6400189
(bug + fix showing that a method's return type is treated as part of the method's type for the purposes of this type erasure)
There is also this request, where someone appears to request the behaviour you describe - only erase the class generic parameters, not other generics - but it was rejected with this reasoning:
The request is to modify type erasure so that in the type declaration
Foo<T>
, erasure only removesT
from parameterized types. Then, it so happens that withinMap<K,V>
's declaration,Set<Map.Entry<K,V>>
erases toSet<Map.Entry>
.But if
Map<K,V>
had a method that took typeMap<String,V>
, its erasure would just beMap<String>
. For type erasure to change the number of type parameters is horrific, especially for compile-time method resolution. We are absolutely not going to accept this request.It is too much to expect to be able to use raw types (
Map
) while still getting some of the type-safety of generics (Set<Map.Entry>
).
With Java Generics, if you use the raw form of a generic class, then all generics on the class, even unrelated generic methods such as your makeSingletonList
and doSomething
methods, become raw. The reason as I understand it is to provide backwards compatibility with Java code written pre-generics.
If there is no use for your generic type parameter T
, then simply remove it from MyGenericClass
, leaving your methods generic with K
. Else you'll have to live with the fact that the class type parameter T
must be given to your class to use generics on anything else in the class.
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