静态变量初始化顺序，Java [英] Order of static variable initialization, Java

问题描述

可能的重复:
Java 静态类初始化
静态块的顺序和类中执行的静态变量?

当我运行这段代码时，答案是 1，我以为是 2.每一步的初始化顺序和k的值是什么?

When I run this code the answer is 1, I thought it would be 2. What is the order of initialization and the value of k in each step?

public class Test {

    static {k = 2;}
    static int k = 1;

    public static void main(String[] args) {
        System.out.println(k);
    }
}

编辑 1:作为k 设置为默认值"的后续内容，为什么下一段代码无法编译?出现错误在定义字段之前无法引用字段".

Edit 1: As a follow up to "k is set to default value" then why this next code doesn't compile? Theres an error "Cannot reference a field before it's defined".

public class Test {

    static {System.out.println(k);}
    static int k=1;

    public static void main(String[] args) {
        System.out.println(k);
    }
}

编辑 2:由于某些我不知道的原因，它 ^ 在而不是k"它的Test.k"时起作用.

Edit 2: For some unknow to me reason it^ works when instead of "k" its "Test.k".

谢谢大家的回答.这就足够了:D

Thanks for all the answers. this will sufice :D

推荐答案

它们按照您编写它们的顺序执行.如果代码是:

They are executed in the order that you write them. If the code is:

public class Test {

    static int k = 1;
    static {k = 2;}

    public static void main(String[] args) {
        System.out.println(k);
    }

}

那么输出变为2.

初始化的顺序是:..the类变量初始化器和类的静态初始化器...，按文本顺序，就好像它们是一个单独的块.

值(对于您的代码)是:k = 0(默认)，然后设置为 2，然后设置回 1.

And the values (for your code) are: k = 0 (default), then it's set to 2, then it's set back to 1.

您可以通过运行以下代码来检查它是否实际设置为 2:

You can check that it's actually set to 2 by running the following code:

private static class Test {

    static {
        System.out.println(Test.k);
        k = 2;
        System.out.println(Test.k);
        }
    static int k = 1;

    public static void main(String[] args) {
        System.out.println(k);
    }
}

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