喷雾：在性状重写功能

Question

我使用的喷雾，有一个基本路线特点，它定义了一些应该实现的功能...

trait ServiceBaseRoute extends HttpService { 
    def function1():Type 
    def function2():Type 

    lazy val serviceBaseRoute = ... 

然后我MIXIN这种特质对其他一些性状...

trait MyRoute1 extends HttpService 
    with ServiceBaseRoute { 
    override def function1():Type = {...} 
    override def function2():Type = {...} 

    val myRoute1 = serviceBaseRoute 

和...

trait MyRoute2 extends HttpService 
    with ServiceBaseRoute { 
    override def function1():Type = {...} 
    override def function2():Type = {...} 

    val myRoute2 = serviceBaseRoute 

最后我作为构建遵循的路线的顶层小号...

trait V1Routes extends HttpService 
    with MyRoute1 
    with MyRoute2 { 

    val v1Routes = 
    pathPrefix("v1") { 
     authenticate(...) { 
     myRoute1 ~ myRoute2 
     } 
    } 

编译罚款等......然而，当我运行这个MyRoute2的功能覆盖覆盖在MyRoute1定义。我在想这是因为我没有MyRoute1和MyRoute2的实际实例，因为它们被简单地混合在一起，并且自MyRoute2之后的MyRoute2的值覆盖了MyRoute1的值之后呢？ 那么实现这种保存每个特征中定义的覆盖的最佳方式是什么？ 我是否需要将MyRoute1和MyRoute2定义为对象？

感谢

Answer 1

根据您的意见，其目的是在MyRoute1和MyRoute2，其中每个被定制serviceBaseRoute创建不同的路线。由于ServiceBaseRoute由两个功能定制，你可以做serviceBaseRoute高阶函数，并通过这两个函数的实现它的子性状：

trait ServiceBaseRoute extends HttpService { 

    def serviceBaseRoute(f1:() => Type, f2:() => Type)= ??? 
} 

trait MyRoute1 extends HttpService 
with ServiceBaseRoute { 

    def f1ImplA(): Type = ??? 

    def f2ImplA(): Type = ??? 

    val myRoute1 = serviceBaseRoute(f1ImplA, f2ImplA) 
} 

trait MyRoute2 extends HttpService 
with ServiceBaseRoute { 

    def f1ImplB(): Type = ??? 

    def f2ImplB(): Type = ??? 

    val myRoute2 = serviceBaseRoute(f1ImplB, f2ImplB) 
} 

trait V1Routes extends HttpService 
with MyRoute1 
with MyRoute2 { 

    val v1Routes = 
    pathPrefix("v1") { 
     authenticate(...) 
     { 
     myRoute1 ~ myRoute2 
     } 
    } 
}