小编Bri*_*ugh的帖子

如何在使用vararg和不使用vararg的方法之间消除Scala中的歧义

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

 public JCommander(Object object)  
 public JCommander(Object object, ResourceBundle bundle, String... args)   
 public JCommander(Object object, String... args)   
Run Code Online (Sandbox Code Playgroud)

我想调用第一个不带 varargs的构造函数.我试过了:

jCommander = new JCommander(cmdLineArgs)
Run Code Online (Sandbox Code Playgroud)

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)
Run Code Online (Sandbox Code Playgroud)

我也尝试使用命名参数,但结果相同:

jCommander = new JCommander(`object` = cmdLineArgs)
Run Code Online (Sandbox Code Playgroud)

我怎么告诉Scala我想调用不带varargs的构造函数?

我正在使用Scala 2.8.0.

java scala scalac

24
推荐指数
3
解决办法
7109
查看次数

标签 统计

java ×1

scala ×1

scalac ×1