为什么类不能使用具有相同签名的方法扩展特征? [英] Why can't a class extend traits with method of the same signature?

问题描述

为什么会出现以下错误?如何解决呢?

Why do I get the error below? How to workaround it?

我假设由于A和B编译为(接口，类)对，因此在编译C时选择正确的静态方法调用即可实现.我希望优先级是按顺序排列的.

I assumed that since A and B compile to (interface,class) pairs, it's a matter of choosing the right static method call to implement when compiling C. I would expect the priority to be according to order.

scala> trait A { def hi = println("A") }
defined trait A

scala> trait B { def hi = println("B") }
defined trait B

scala> class C extends B with A
<console>:6: error: error overriding method hi in trait B of type => Unit;
 method hi in trait A of type => Unit needs `override' modifier
       class C extends B with A

scala> trait A { override def hi = println("A") }
<console>:4: error: method hi overrides nothing
       trait A {override def hi = println("A")}

请注意，在Ruby中效果很好:

Note that in Ruby this works well:

>> module B; def hi; puts 'B'; end; end
=> nil
>> module A; def hi; puts 'A'; end; end
=> nil
>> class C; include A; include B; end
=> C
>> c = C.new
=> #<C:0xb7c51068>
>> c.hi
B
=> nil

推荐答案

这在2.8和2.11中对我有效，并且允许您对特征A或B不打扰:

This works for me in 2.8 and 2.11, and would allow you to be non-intrusive in traits A or B:

trait A { def hi = println("A") }
trait B { def hi = println("B") }

class C extends A with B {
  override def hi = super[B].hi
  def howdy = super[A].hi // if you still want A#hi available
}

object App extends Application {
  (new C).hi // prints "B"
}

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