java 为什么在这种情况下会生成 classname$1.class？
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Why does classname$1.class generate in this situation?
提问by handrenliang
I wrote the following code to implement the Singleton pattern:
我编写了以下代码来实现单例模式：
public final class Test {
static final class TestHolder {
private static final Test INSTANCE = new Test();
}
private Test() {}
public static Test getInstance() {
return TestHolder.INSTANCE;
}
}
When I compile this file, it should generate Test.class and Test$TestHolder.class, but it also generates Test$1.class. This doesn't make sense. So why and how would this be?
当我编译这个文件时，它应该生成Test.class 和Test$TestHolder.class，但它也会生成Test$1.class。这没有意义。那么为什么会这样？
回答by Ernest Friedman-Hill
Class TestHolderneeds to call the private constructor in Test. But it's private, and can't actually be called from another class. So the compiler plays a trick. It adds a new non-private constructor to Testwhich only it knows about!That constructor takes an (unused) instance of this anonymous class Test$1-- which nobody knows exists. Then TestHoldercreates an instance of Test$1and calls thatconstructor, which is accessible (it's default-protected.)
类TestHolder需要调用Test. 但它是私有的，实际上不能从另一个类调用。所以编译器玩了一个把戏。它添加了一个Test只有它知道的新的非私有构造函数！该构造函数采用此匿名类的（未使用的）实例Test$1——没人知道它存在。然后TestHolder创建一个实例Test$1并调用该构造函数，它是可访问的（它是默认保护的。）
You can use javap -c Test(and javap -c Test\$1, and javap -c Test\$TestHolder) to see the code. It's quite clever, actually!
您可以使用javap -c Test(and javap -c Test\$1, and javap -c Test\$TestHolder) 查看代码。其实还挺聪明的！
