java - 为什么 super 方法不可见/无法解决？

Question

interface Problematic {
    void method();
}
class Parent {
    void method(int arg) { }
}

class Child extends Parent {
    void test() {
        new Problematic() {
            @Override public void method() { 
                // javac error: method method in class <anonymous Problematic> cannot be applied to given types;
                // required: no arguments, found: int
                method(0);
                Child.this.method(0); // works just fine
                Child.super.method(0); // works just fine
            }
        };
    }
}

IntelliJ IDEA 也给出警告:

Method 'method()' recurses infinitely, and can only end by throwing an exception

Answer 1

From the Java Language Specification,

If the form is MethodName, that is, just an Identifier, then:

If the Identifier appears in the scope of a visible method declaration with that name (§6.3, §6.4.1), then:

• If there is an enclosing type declaration of which that method is a member, let T be the innermost such type declaration. The class or interface to search is T.

• This search policy is called the "comb rule". It effectively looks for methods in a nested class's superclass hierarchy before looking for methods in an enclosing class and its superclass hierarchy. See §6.5.7.1 for an example.

• Otherwise, the visible method declaration may be in scope due to one or more single-static-import or static-import-on-demand declarations. There is no class or interface to search, as the method to be invoked is determined later (§15.12.2.1).

您正在调用匿名内部类中的方法，该类是 Problematic 的子类型。这使得 Child 成为它的封闭类型。在您的情况下， T 是匿名内部类，因为它是最里面的类型。因此，搜索方法的类是 T，即。 Problematic 的匿名子类。

在later the resolution of method invocation expressions的步骤, 决定方法 Problematic#method() 没有参数，因为它没有声明任何参数。因此它不适用并会引发编译器错误。