为什么此Java代码会产生堆栈溢出错误? [英] Why does this Java code create a stack overflow error?

问题描述

下面的代码在执行时会产生堆栈溢出错误.但是，如果删除

The code below produces a stack over flow error when it is executed. However If remove either of

• 静态最终GenerateStackOverflow E1 =新的GenerateStackOverflow("value1");
• final GenerateStackOverflow E2 =新的GenerateStackOverflow("value2");

它运行时没有堆栈溢出错误.如果我有以上两行，但是如果只有这两行在类中，那么我怎么会出现堆栈溢出错误呢?

It runs without a stack over flow error. How come I get a stack overflow error if I have the above two lines but no error if only one of the lines is in the class?

public class GenerateStackOverflow {

    private final String value; 

    static final GenerateStackOverflow E1 = new GenerateStackOverflow("value1");
    final GenerateStackOverflow E2 = new GenerateStackOverflow("value2");


    public GenerateStackOverflow(String value) {
        System.out.println("GenerateStackOverflow.GenerateStackOverflow()");
        this.value = value;
    }

    public String getValue() {
        return value;
    }

    public static void main(String[] args) {
        GenerateStackOverflow.class.getName();
    }
}

推荐答案

两个都需要生成 StackOverflowError .当您添加以下行时:

Both are needed to generate a StackOverflowError. When you include this line:

static final GenerateStackOverflow E1 = new GenerateStackOverflow("value1");

首次访问该类时，会创建 GenerateStackOverflow 的实例.

An instance of GenerateStackOverflow is created when the class if first accessed.

不包括此行:

final GenerateStackOverflow E2 = new GenerateStackOverflow("value2");

一切都很好.但是这条线很关键.每次创建 GenerateStackOverflow 的实例时，它都会尝试初始化其成员变量 E2 -另一个 GenerateStackOverflow 对象.然后那个实例将把其 E2 初始化为另一个 GenerateStackOverflow 对象.这一直持续到发生 StackOverflowError .

things are fine. But this line is critical. Each time an instance of GenerateStackOverflow is created, it attempts to initialize its member variable E2 -- another GenerateStackOverflow object. Then that instance will have its E2 initialized to another GenerateStackOverflow object. This continues until a StackOverflowError occurs.

如果包括第二行但不包括第二行，则不会创建任何实例，并且永远不会输入无限递归.

If the second line is included but the first isn't, then no instance is created and the infinite recursion is never entered.

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