为什么在没有任何参数的情况下调用一个接受Any的方法是合法的？

Question

更别说为什么它是合法的，为什么这甚至会回归

scala> class Bla { 
    def hello(x: Any): Boolean = x.toString.length == 2 
}
defined class Bla

scala> new Bla().hello() 
res0: Boolean = true 

Answer 1

使用-deprecation运行提供此

scala> scala> class Bla { 
  def hello(x: Any): Boolean = x.toString.length == 2 
}
defined class Bla

scala> new Bla().hello()
<console>:13: warning: Adaptation of argument list by inserting () has been deprecated: leaky (Object-receiving) target makes this especially dangerous.
        signature: Bla.hello(x: Any): Boolean
  given arguments: <none>
 after adaptation: Bla.hello((): Unit)
       new Bla().hello()
                      ^
res0: Boolean = true

警告信息的含义是：

hello()被解释为hello(()) 从().toString = "()"开始，该方法返回true。