问题描述
为什么从Java使用所有Scala vararg方法时，它们似乎都接受Seq变量，因此不能用作Java本机vararg方法.这是一个错误吗?
Why do all scala vararg methods, when used from java, seem to accept a Seq of variables, and can't be used as java native vararg methods. Is this a bug?
例如，Buffer具有方法def append(elems: A*): Unit.但是在Java中，它还有另一个签名:void append(Seq<A>).
For instance, Buffer has method def append(elems: A*): Unit. But in java it has another signature: void append(Seq<A>).
推荐答案
这不是错误.这是一种设计选择，它支持在Scala中使用vararg而不是与Java进行互操作.例如，它允许您将List传递到Scala varargs方法中，而不必在途中将其转换为Array.
It is not a bug. It is a design choice that favors vararg use within Scala over interoperability with Java. For example, it allows you to pass a List into a Scala varargs method without having to convert it to an Array on the way.
如果您需要使用Java的Scala变量，则应该创建一些scala Seq.例如，您可以编写Java包装程序来自动创建一个数组，然后使用Predef对象中的genericWrapArray方法.
If you need to use Scala varargs from Java, you should create some scala Seq instead. You can, for example, write a Java wrapper to get an array automatically created, and then use the genericWrapArray method from the Predef object.
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