问题描述
为什么 File :: isDirectory 在以下示例中作为FileFilter正常工作？
Why File::isDirectory works fine as a FileFilter in the below example?
File[] files = new File(".").listFiles(File::isDirectory);
listFiles方法需要FileFilter作为参数
The listFiles method requires a FileFilter as a parameter
public File[] listFiles(FileFilter filter) {
...
}
FileFilter是一个功能接口，它有一个方法 accept ，带有File参数
The FileFilter is a functional interface which has one method accept with a File parameter
boolean accept(File pathname);
File类中的isDirectory方法没有参数
And isDirectory method from the File class has no parameters
public boolean isDirectory() {
...
}
推荐答案
为了使事情更清楚，方法引用 File :: isDirectory 等同于以下lambda ：
To make things clearer the method reference File::isDirectory is equivalent to the following lambda:
file -> file.isDirectory()
如你所见我们传递的是文件参数然后在它上面调用 isDirectory 返回 boolean 因此满足 FileFilter 界面中的SAM。
As you can see we're passing a File parameter then calling isDirectory upon it which returns boolean hence satisfies the SAM in the FileFilter interface.
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