如何在 Scala 中为类实例生成唯一 ID？
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How to generate an unique ID for an class instance in Scala?
提问by Gregosaurus
I have a class that needs to write to a file to interface with some legacy C++ application. Since it will be instantiated several times in a concurrent manner, it is a good idea to give the file an unique name.
我有一个类需要写入文件以与某些遗留 C++ 应用程序交互。由于它将以并发方式多次实例化，因此给文件一个唯一的名称是个好主意。
I could use System.currentTimemili or hashcode, but there exists the possibility of collisions.
Another solution is to put a varfield inside a companion object.
我可以使用 System.currentTimemili 或 hashcode，但存在冲突的可能性。另一种解决方案是将一个var字段放在伴随对象中。
As an example, the code below shows one such class with the last solution, but I am not sure it is the best way to do it (at least it seems thread-safe):
例如，下面的代码显示了一个这样的类和最后一个解决方案，但我不确定这是最好的方法（至少它看起来是线程安全的）：
case class Test(id:Int, data: Seq[Double]) {
//several methods writing files...
}
object Test {
var counter = 0
def new_Test(data: Seq[Double]) = {
counter += 1
new Test(counter, data)
}
}
采纳答案by senia
it is a good idea to give the file an unique name
给文件一个唯一的名字是个好主意
Since all you want is a file, not id, the best solution is to create a file with unique name, not a class with unique id.
由于您想要的只是一个文件，而不是 id，因此最好的解决方案是创建一个具有唯一名称的文件，而不是具有唯一 id 的类。
You could use File.createTempFile:
你可以使用File.createTempFile：
val uniqFile = File.createTempFile("myFile", ".txt", "/home/user/my_dir")
Vladimir Matveevmentionedthat there is a better solution in Java 7 and later - Paths.createTempFile:
Vladimir Matveev提到在 Java 7 及更高版本中有更好的解决方案 - Paths.createTempFile：
val uniqPath = Paths.createTempFile(Paths.get("/home/user/my_dir"), "myFile", ".txt"),
回答by Gregosaurus
Did you try this :
你试过这个吗：
def uuid = java.util.UUID.randomUUID.toString
See UUIDjavadoc, and also How unique is UUID?for a discussion of uniqueness guarantee.
请参阅UUIDjavadoc，以及UUID 有多独特？讨论唯一性保证。
