NullPointerException

本文介绍了NullPointerException | enum构造函数中的`this`导致NPE的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 public class Test { public static void main(String[] args) { Platform1 p1=Platform1.FACEBOOK; //giving NullPointerException. Platform2 p2=Platform2.FACEBOOK; //NO NPE why? } }

enum Platform1{ FACEBOOK,YOUTUBE,INSTAGRAM; Platform1(){ initialize(this); }; public void initialize(Platform1 platform){ switch (platform) { //platform is not constructed yet,so getting `NPE`. //ie. we doing something like -> switch (null) causing NPE.Fine! case FACEBOOK: System.out.println("THIS IS FACEBOOK"); break; default: break; } } }

enum Platform2{ FACEBOOK("fb"),YOUTUBE("yt"),INSTAGRAM("ig"); private String displayName; Platform2(String displayName){ this.displayName=displayName; initialize(this); }; public void initialize(Platform2 platform){ switch (platform.displayName) { //platform not constructed,even No `NPE` & able to access its properties. //switch (null.displayName) -> No Exception Why? case "fb": System.out.println("THIS IS FACEBOOK"); break; default: break; } } }

任何人都可以解释为什么那里是 NullPointerException 在 Platform1 中，但不在 Platform2 中。在第二种情况下，我们如何能够访问枚举对象及其属性，甚至在构造对象之前？

Can anyone explain me why there is NullPointerException in Platform1 but not in Platform2. How in the second case we are able to access the enum object and its properties, even before the object is constructed?

推荐答案

确切地说。正如@PeterS在正确构造之前提到的那样使用枚举导致NPE，因为在未构造的枚举上调用了values（）方法。

Exactly. Just as @PeterS mentioned using enum before it has been properly constructed is causing NPE, because values() method is being called on un-constructed enum.

还有一点，我想在此处添加 Platform1 和 Platform2 两者都试图在switch（）中使用未构造的枚举，但是NPE是仅在 Platform1 中。这背后的原因如下： -

One more point, I would like to add here that Platform1 and Platform2 both are trying to use unconstructed enum in switch() but NPE is only in Platform1. Reason behind this is as follows :-

public void initialize(Platform1 platform){ switch (platform) {

以上来自 Platform1 枚举的代码正在使用 platform 在switch中的enum对象，其中使用内部 $ SwitchMap $ Platform1 [] 数组并初始化此数组 values（）使用方法，因此您获得NPE。但是在 Platform2 中，开关（platform.displayName）是 displayName 已经初始化并且发生了字符串比较，因此没有NPE。

Above piece of code from Platform1 enum is using platform enum object in switch where internally $SwitchMap$Platform1[] array is used and to initialize this array values() method is utilized, thus you get NPE. But in Platform2, switch (platform.displayName) is comparison on displayName which is already initialized and a string comparison occurs thus no NPE.

以下是反编译代码的片段： -

Following are fragments of decompiled code :-

Platform1

Platform1

static final int $SwitchMap$Platform1[] = new int[Platform1.values().length];

Platform2

Platform2

switch ((str = platform.displayName).hashCode()) { case 3260: if (str.equals("fb")) {