静态变量与非静态变量 [英] static variable vs non static variable

问题描述

我定义了一个对象并声明了一个静态变量 i 。在 get（）方法中，当我尝试打印实例和类变量时，两者都打印相同的值。

I have defined an object and declared a static variable i. In the get() method, when I try to print the instance and class variable, both print the same value.

不是 this.i 一个实例变量？它应该打印0而不是50吗？

Isn't this.i an instance variable? Should it print 0 instead of 50?

public class test {
    static int i = 50;
    void get(){
        System.out.println("Value of i = " + this.i);
        System.out.println("Value of static i = " + test.i);
    }

    public static void main(String[] args){
        new test().get();
    }

}

推荐答案

不，只有一个变量 - 你没有声明任何实例变量。

No, there's only one variable - you haven't declared any instance variables.

不幸的是，Java允许你访问静态成员，就像你通过一个访问它一样相关类型的参考。这是一个设计缺陷IMO，一些IDE（例如Eclipse）允许您将其标记为警告或错误 - 但它是语言的一部分。您的代码是有效的：

Unfortunately, Java lets you access static members as if you were accessing it via a reference of the relevant type. It's a design flaw IMO, and some IDEs (e.g. Eclipse) allow you to flag it as a warning or an error - but it's part of the language. Your code is effectively:

System.out.println("Value of i = " + test.i);
System.out.println("Value of static i = " + test.i);

如果你做通过相关类型的表达式，它不会甚至检查值 - 例如：

If you do go via an expression of the relevant type, it doesn't even check the value - for example:

test ignored = null;
System.out.println(ignored.i); // Still works! No exception

但仍会评估任何副作用。例如：

Any side effects are still evaluated though. For example:

// This will still call the constructor, even though the result is ignored.
System.out.println(new test().i);

这篇关于静态变量与非静态变量的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！