覆盖接口的变量？ [英] Overriding interface's variable?

问题描述

当我从各种Java书籍和教程中读到时，在接口中声明的变量是常量，不能被覆盖。

我做了一个简单的代码来测试它

  interface A_INTERFACE 
 {
 int var = 100; 
} 
 
类A_CLASS实现A_INTERFACE 
 {
 int var = 99; 
 // test 
 void printx（）
 {
 System.out.println（var =+ var）; 
} 
} 
 
 class hello 
 {
 
 public static void main（String [] args）
 {
 new A_CLASS（）。printx（）; 
} 
} 
  

并打印出
var = 99

var是否被覆盖？我完全糊涂了。感谢您的任何建议！

---

非常感谢大家！我对这个界面很新。 影子是理解这一点的关键词。我查阅相关材料并立即了解它。

解决方案

它没有被覆盖，但被遮蔽，因为常数会产生额外的混乱在界面中也是静态的。

试试这个：

  A_INTERFACE o = new A_CLASS（）; 
 System.out.println（o.var）; 
  

你应该得到一个关于以非静态方式访问静态字段的编译时警告。 / p>

现在这个

  A_CLASS o = new A_CLASS（）; 
 System.out.println（o.var）; 
 System.out.println（A_INTERFACE.var）; //坏名字，顺便说一句，因为它是const 
  

As I read from various Java book and tutorials, variables declared in a interface are constants and can't be overridden.

I made a simple code to test it

interface A_INTERFACE
{ 
    int var=100; 
}

class A_CLASS implements A_INTERFACE
{ 
    int var=99; 
    //test
    void printx()
    {
        System.out.println("var = " + var);
    }
}

class hello
{

    public static void main(String[] args)
    {
        new A_CLASS().printx();
    }
}

and it prints out var = 99

Is var get overridden? I am totally confused. Thank you for any suggestions!

---

Thank you very much everyone! I am pretty new to this interface thing. "Shadow" is the key word to understand this. I look up the related materials and understand it now.

解决方案

It is not overridden, but shadowed, with additional confusion because the constant in the interface is also static.

Try this:

A_INTERFACE o = new A_CLASS();
System.out.println(o.var);

You should get a compile-time warning about accessing a static field in a non-static way.

And now this

A_CLASS o = new A_CLASS();
System.out.println(o.var);
System.out.println(A_INTERFACE.var);  // bad name, btw since it is const

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