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问题描述
当我编译这段代码时:
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方法是如何被认为是public的?
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的public
.并且接口中声明的所有变量都是隐式的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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