Scala 如何在具有隐式参数的函数上调用Private

如何为这些类调用XYZ.doSomething方法：

XYZ.scala

object XYZ {

  private def doSomething(i : Int)(implicit s: String): String {
    s + i.toString
  }
}

XYZTest.scala

class XYZTest extends FunSpec with PrivateMethodTester {
  describe("SomeTest") {
    it("Can't find a private method named: doSomething :( ") {
      implicit lazy val someStr: String = "sahil"
      val doSomething = PrivateMethod[String]('doSomething)
      val myStr = XYZ invokePrivate doSomething(1)
      assert(myStr == "sahil1")
    }
  }
  describe("SomeTest") {
    it("This doesn't even compile :( ") {
      val doSomething = PrivateMethod[String]('doSomething)
      val myStr = XYZ invokePrivate doSomething(1)("sahil")
      assert(myStr == "sahil1")
    }
  }
}

正确答案是：

object XYZ {

  private def doSomething(i : Int)(implicit s: String): String = {
    s + i.toString
  }
}

class XYZTest extends FunSpec with PrivateMethodTester {
  describe("SomeTest") {
    it("Can't find a private method named: doSomething :( ") {
      implicit lazy val someStr: String = "sahil"
      val doSomething = PrivateMethod[String]('doSomething)
      val myStr = XYZ invokePrivate doSomething(1, someStr)
      assert(myStr == "sahil1")
    }
  }
}

invokePrivate

方法的简要视图显示，它不支持

implicits

，但看起来它将

currying

参数列表视为参数的一般序列

def invokePrivate[T]（调用：PrivateMethodTester.this.invoke[T]）

其中，

args

是顺序：

最终类调用[T]（val-methodName:scala.Symbol，val-args:scala.Any*）