通用類型自引用類型的Java子類問題

Question

import java.util.*; 

// Let's define a self-referential type: 
class SelfReferential<T extends SelfReferential<T>> {} 

//A complete (i.e. not parameterized) subtype of SelfReferential: 
class SubclassA extends SelfReferential<SubclassA> {} 

//A partial (i.e. parameterized) subtype of SelfReferential: 
class SubclassB<T extends SubclassB<T>> extends SelfReferential<T> {} 

//Two complete subtypes of SubclassB 
class SubclassB1 extends SubclassB<SubclassB1> {}  
class SubclassB2 extends SubclassB<SubclassB2> {} 

//Now let's define a generic type over SelfReferential: 
class Generic<T extends SelfReferential<T>> {} 

//No problem creating a subtype for A, B1 or B2 
class GenericA extends Generic<SubclassA> {} 
class GenericB1 extends Generic<SubclassB1> {}  
class GenericB2 extends Generic<SubclassB2> {} 

//We can also defined a parameterize type for specific types extending SubclassB 
class GenericB<T extends SubclassB<T>> extends Generic<T> {} 

//However, it does not seem possible to define a non-parameterized subtype of Generic of ANY subtype of SublassB 
//My goal is to provide a type alias for GenericB<? extends SubclassB<?>> to avoid 
//having to mention it everywhere in the code. This is like providing an alias for ArrayList<String> using 
class ArrayListOfString extends ArrayList<String> {} 

//Unsucessful attempts: 
//class GenericAnyB extends Generic<SubclassB> {} //ERROR: bound mismatch 
//class GenericAnyB extends Generic<SubclassB<?>> {} //ERROR: bound mismatch 
//class GenericAnyB extends Generic<? extends SubclassB<?>> {} //ERROR: invalid syntax: a supertype cannot specify any wildcard 
//class GenericAnyB extends Generic<SubclassB<? extends SubclassB>> {} //ERROR: bound mismatch 
//class GenericAnyB extends Generic<SubclassB<SubclassB<SubclassB<SubclassB<SubclassB<SubclassB>>>>>> {} // well... 
//class GenericAnyB extends <T extends SubclassB<T>> Generic<T> {} //ERROR: this syntax is illegal 

底線，我無法在extends子句中指定「引用循環」。

問題：這是Java語言的限制嗎？

Answer 1

你說得對，這是不可能的，就像使用自引用類型聲明一個變量是不可能的，沒有通配符或原始類型是不可能的。您將無法直接實例化SubclassB，原因與您不能將其用作沒有自引用類型參數的綁定的原因相同。

看到這個職位的詳細討論這一限制：Self bound generic type with fluent interface and inheritance

的底線是GenericAnyB將需要的通用本身使用SubclassB爲綁定：

class GenericAnyB<T extends SubclassB<T>> extends Generic<T> { } 

這只是增加了一個額外的在任何事物可用之前進入層級：

class GenericB1 extends GenericAnyB<SubclassB1> { } 
class GenericB2 extends GenericAnyB<SubclassB2> { }