As a couple of other answers have explained, the Java compiler takes a file name as an argument, whereas the interpreter takes a class name.

So you give the .java extension to the compiler because it's part of the file name, but you don't give it to the interpreter because it's not part of the class name.

But then, you might wonder, why didn't they just design the Java interpreter differently so that it would take a file name? The answer to that is that classes are not always loaded from .class files. Sometimes they come from JAR archives, sometimes they come from the internet, sometimes they are constructed on the fly by a program, and so on. A class could come from any source that can provide the binary data needed to define it. Perhaps the same class could have different implementations from different sources, for example a program might try to load the most up-to-date version of some class from a URL, but would fall back to a local file if that failed. The designers of Java thought it best that when you're trying to run a program, you don't have to worry about having to track down the source that defines the class you're running. You just give the fully qualified class name and let Java (or rather, its ClassLoaders) do the hard work of finding it.

正如其他几个答案所解释的那样，Java 编译器采用文件名作为参数，而解释器采用类名。因此，您将 .java 扩展名提供给编译器，因为它是文件名的一部分，但您没有将其提供给解释器，因为它不是类名的一部分。
但是，您可能想知道，为什么他们不设计不同的 Java 解释器，以便它采用文件名？答案是类并不总是从 .class 文件加载。有时它们来自 JAR 包，有时它们来自互联网，有时它们是由程序动态构建的，等等。一个类可以来自任何可以提供定义它所需的二进制数据的来源。也许同一个类可能有来自不同来源的不同实现，例如，程序可能会尝试从 URL 加载某个类的最新版本，但如果失败，则会回退到本地文件。 Java 的设计者认为，当您尝试运行程序时，最好不必担心必须追踪定义您正在运行的类的源代码。您只需给出完全限定的类名，然后让 Java（或者更确切地说，它的 ClassLoader）来完成查找它的艰苦工作。