无法从静态上下文引用非静态类 [英] non-static class cannot be referenced from a static context

问题描述

可能重复：

为什么我得到非静态变量，这不能从静态上下文中引用？

以下是代码

 公共类Stack 
 {
私有类节点{
 ... 
} 
 ... 
 public static void main （String [] args）{
 Node node = new Node（）; //生成编译错误
} 
} 
  

错误说：

非静态类节点无法从静态上下文引用

为什么我不应该在main（）方法中引用Node类？

解决方案

Java中的非静态嵌套类包含对父类的实例的隐式引用。因此，要实例化节点，您还需要一个 Stack 的实例。在静态上下文（main方法）中，没有要引用的 Stack 的实例。因此，编译器指示它不能构造节点。

如果你使 Node 一个静态类（或常规外部类），然后它不需要对 Stack 的引用，并且可以直接在静态main方法中实例化。 / p>

请参阅 Java语言规范，第8章了解详细信息，例如例8.1.3-2。

Possible Duplicate:
Why do I get "non-static variable this cannot be referenced from a static context"?

Here are the codes

public class Stack
{
    private class Node{
        ...
    }
    ...
    public static void main(String[] args){
         Node node = new Node(); // generates a compiling error
    }
}  

the error says:

non-static class Node cannot be referenced from a static context

Why shouldn't I refer the Node class in my main() method ?

解决方案

A non-static nested class in Java contains an implicit reference to an instance of the parent class. Thus to instantiate a Node, you would need to also have an instance of Stack. In a static context (the main method), there is no instance of Stack to refer to. Thus the compiler indicates it can not construct a Node.

If you make Node a static class (or regular outer class), then it will not need a reference to Stack and can be instantiated directly in the static main method.

See the Java Language Specification, Chapter 8 for details, such as Example 8.1.3-2.

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