public static void main（）访问非静态变量

问题描述

它说非静态变量不能在静态方法中使用。但是public static void main呢。怎么样？

Its said that non-static variables cannot be used in a static method.But public static void main does.How is that?

推荐答案

不，它没有。

public class A {
  int a = 2;
  public static void main(String[] args) {
    System.out.println(a); // won't compile!!
  }
}

但是

public class A {
  static int a = 2;
  public static void main(String[] args) {
    System.out.println(a); // this works!
  }
}

或者如果你实例化 A

public class A {
  int a = 2;
  public static void main(String[] args) {
    A myA = new A();
    System.out.println(myA.a); // this works too!
  }
}

此外

public class A {
  public static void main(String[] args) {
    int a = 2;
    System.out.println(a); // this works too!
  }
}

将起作用，因为 a 是一个局部变量，而不是实例变量。无论方法是否为 static ，方法局部变量在执行方法期间始终可以访问。

will work, since a is a local variable here, and not an instance variable. A method local variable is always reachable during the execution of the method, regardless of if the method is static or not.

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