java - 方法的动态调度和访问级别

Question

这个问题已经有答案了:

Private methods in Inheritance (9 个回答)

已关闭 9 年前。

考虑以下类:

public class A {
    public void foo() {
        System.out.println("A.foo()");
    }

    public void bar() {
        System.out.println("A.bar()");
        foo();
    }
}

public class B extends A {
    public void foo() {
        System.out.println("B.foo()");
    }

    public static void main(String[]
    args) {
        A a = new B();
        a.bar();
    }
}

这段代码的输出是 A.bar()然后B.foo() 。我注意到，如果我将方法 foo() 的访问级别从 public 更改为至private输出是:A.bar()然后A.foo() .

为什么？

Answer 1

如果A.foo()是私有(private)的，那么它不能被子类覆盖——任何其他类基本上应该不知道私有(private)成员的存在。您无法覆盖您无法“看到”的成员。

来自section 8.4.8.1 of the JLS :

An instance method m_C declared in or inherited by class C, overrides from C another method m_A declared in class A, iff all of the following are true:

• ...

• One of the following is true:

  • m_A is public.
  • m_A is protected.
  • m_A is declared with package access in the same package as C, and either C declares m_C or m_A is a member of the direct superclass of C.
  • m_A is declared with package access and m_C overrides m_A from some superclass of C.
  • m_A is declared with package access and m_C overrides a method m' from C (m' distinct from m_C and m_A), such that m' overrides m_A from some superclass of C.