如何在使用vararg和不使用vararg的方法之间消除Scala中的歧义 [英] How do I disambiguate in Scala between methods with vararg and without

问题描述

我正在尝试使用Scala中的java jcommander库。 java JCommander类有多个构造函数：

I'm trying to use the java jcommander library from Scala. The java JCommander class has multiple constructors:

 public JCommander(Object object)  
 public JCommander(Object object, ResourceBundle bundle, String... args)   
 public JCommander(Object object, String... args)   

我想调用第一个带 no varargs的构造函数。我试过了：

I want to to call the first constructor that takes no varargs. I tried:

jCommander = new JCommander(cmdLineArgs)

我收到错误：

error: ambiguous reference to overloaded definition,
both constructor JCommander in class JCommander of type (x$1: Any,x$2: <repeated...>[java.lang.String])com.beust.jcommander.JCommander
and  constructor JCommander in class JCommander of type (x$1: Any)com.beust.jcommander.JCommander
match argument types (com.lasic.CommandLineArgs) and expected result type com.beust.jcommander.JCommander
jCommander = new JCommander(cmdLineArgs)

我也尝试过使用命名参数，但结果相同：

I've also tried using a named parameter, but got the same result:

jCommander = new JCommander(`object` = cmdLineArgs)

如何告诉Scala我想调用不带varargs的构造函数？

How do I tell Scala I want to call the constructor that doesn't take varargs?

我正在使用Scala 2.8.0。

I'm using Scala 2.8.0.

推荐答案

抱歉，我现在意识到这是Java的已知互操作性问题。请参阅此问题和门票。我所知道的唯一工作就是创建一个小的Java类来消除这些调用的歧义。

Sorry, I now realize this is a known interoperability problem with Java. See this question and the ticket. The only work around I know of is to create a small Java class just to disambiguate these calls.

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