Java因实现访问较弱的接口方法而出错 [英] Java getting an error for implementing interface method with weaker access

问题描述

编译此代码时：

interface Rideable {
    String getGait();
}

public class Camel implements Rideable {
    int x = 2;

    public static void main(String[] args) {
        new Camel().go(8);
    }

    void go(int speed) {
        System.out.println((++speed * x++) 
        + this.getGait());
    }

    String getGait() {
        return " mph, lope";
    }
}

我收到以下错误：

Camel.java:13: error: getGait() in Camel cannot implement getGait() in Rideable
String getGait() {
       ^
  attempting to assign weaker access privileges; was public
1 error

接口中声明的getGait方法如何被公开？

How is the getGait method declared in the interface considered public?

推荐答案

在接口内声明的方法是隐式 public 。并且在接口中声明的所有变量都是隐式 public static final （常量）。

Methods declared inside interface are implicitly public. And all variables declared in the interface are implicitly public static final (constants).

public String getGait() {
  return " mph, lope";
}

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