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在Java,Can t 进入保护下级超级成员
原标题:In Java , Can t access protected members of super-class from sub-class
class Super {
   protected int a;
   protected Super(int a) { this.a = a; }
}

class Sub extends Super {
    public Sub(int a) { super(a); }
    public Sub() { this.a = 5; }
}




  public Sub() { this.a = 5; }  

a.a=5=t工作。 为什么如此? 应当继承受保护和公开的成员。

最佳回答

问题在于,你无法利用变数,但你没有叫作基底构造:

class Super {
   protected int a;
   protected Super(int a) { this.a = a; }
}

class Sub extends Super {
    public Sub(int a) { super(a); }
    public Sub() { 
       super(0);  // <-- call base constructor
       this.a = 5; 
    }
}

这是因为你没有为<代码>Super界定违约构造者,因此,衍生的班子不知道,如果你不指明哪位构造者,会叫谁。

问题回答

您在<代码>Sub中的无参数构造者试图在Super中暗指无参数的构造,而这种构造确实存在,这就是(我假定)为什么你重犯了汇编错误。

哪些工作不可行? 页: 1

您没有违约构造——见public Sub(){this.a = 5;}。 如果我编辑你的法典,我会:

cannot find symbol constructor Super()

因此,你们或是必须有一个缺省建筑商或做以下工作:





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