Question

我有scala函数，如下所示，

scala> def getOrders: (String, String) => Seq[String] = (user: String, apiToken: String) => Seq.empty[String] 
def getOrders: (String, String) => Seq[String]

scala> getOrders("prayagupd", "A1B2C3") 
val res0: Seq[String] = List()

我想传递第三个参数作为隐式参数，但对于函数来说似乎不可能。

这是我想要的方法，

scala> def getOrders(user: String, apiToken: String)(implicit clientType: String) = Seq.empty[String] 
def getOrders
  (user: String, apiToken: String)(implicit clientType: String): Seq[String]

scala> implicit val clientType: String = "android" 
implicit val clientType: String = "android"

scala> getOrders("prayagupd", "A1B2C3") 
val res2: Seq[String] = List()

由于apply函数是预定义的，因此无法获得额外的接受隐式参数。

scala> new Function2[String, String, Seq[String]] {
          def apply(user: String, apiToken: String): Seq[String] = Seq.empty
         } 
val res4: (String, String) => Seq[String] = <function2>

重载也不起作用，

scala> new Function2[String, String, Seq[String]] {
          def apply(user: String, apiToken: String): Seq[String] = Seq.empty
          def apply(user: String, apiToken: String)(implicit clientType: String) = Seq("order1")
         } 
val res9: (String, String) => Seq[String] = <function2>

scala> implicit val clientType: String = "device" 
implicit val clientType: String = "device"

scala> res9("prayagupd", "apiToken") 
val res10: Seq[String] = List()

是不是implicit根本不建议用于功能，或者我错过了什么？

Answer 1

实验，你的函数可能表达如下，没有隐含的：

scala> def getOrders: (String, String) => (String) => Seq[String] = (user: String, apiToken: String) => (clientType: String) => Seq.empty[String]
def getOrders: (String, String) => String => Seq[String]

四处寻找......它不像implicit那样可能会让你想要你想要的东西。

对related question的回答表明原因：getOrders＆＃34; ...是方法，而不是函数，并且不会尝试eta-expansion（将方法转换为函数）直到隐式申请之后。＆＃34;似乎implicits在方法级别解决，而不是在函数级别解决。

具有隐式参数的scala函数

1 个答案: