java - 带有接口(interface)的 "Cannot reduce the visibility of the inherited method"的含义

Question

我有两个文件:

public interface PrintService {
    void print(PrintDetails details);
    class PrintDetails {
        private String printTemplate;
    }
    public interface Task {
        String ACTION = "print";
    }
}

和

public class A implements PrintService {
    void print(PrintDetails details) {
        System.out.println("printing: " + details);
    }
    String action = PrintService.Task.ACTION;   
}

我认为代码看起来没问题，但我在第二个文件中收到错误，行 void print(PrintDetails details) { 指出:

Cannot reduce the visibility of the inherited method from PrintService.

谁能解释一下这对我意味着什么？

Answer 1

在 Java 接口(interface)中，每个方法默认都是public:

Every method declaration in the body of an interface is implicitly abstract, so its body is always represented by a semicolon, not a block.

Every method declaration in the body of an interface is implicitly public. [..]

在实现类中，不允许通过不指定访问修饰符来降低可见性:

void print(){..}

您正在指定访问级别 default ，其可见性低于 public。