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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