是否有一种单行方式来提供对一个值的命名using
引用,该值在没有即将被弃用的implicit
关键字的情况下隐式可用(即通过语法可用)?根据文档,我希望以下工作(在 SBT 下,scalaVersion := "3.0.0-M2"
:
trait Greeter {
def sayHello(username: String): Unit
def shutdown(): Unit
}
def greet(username: String)(using g: Greeter): Unit = {
g.sayHello(username)
}
object Foo extends App {
given greeter: Greeter = new Greeter {
override def sayHello(username: String): Unit = println(s"Hello, ${username}!")
override def shutdown(): Unit = println("Shutting down")
}
try {
greet("world")
} finally {
greeter.shutdown()
}
}
但这失败了
[error] -- Error: /path/to/project/src/main/scala/Foo.scala:12:15
[error] 12 | given greeter: Greeter = new Greeter {
[error] | ^
[error] | end of statement expected but ':' found
[error] -- [E040] Syntax Error: /path/to/project/src/main/scala/Foo.scala:12:17
[error] 12 | given greeter: Greeter = new Greeter {
[error] | ^^^^^^^
[error] | ';' expected, but identifier found
现在:我可以通过多种方式解决这个问题,但要么文档令人困惑(或错误),要么我误解了一些相当基本的东西。
解决方法 1(如此处建议):
lazy val greeter: Greeter = new Greeter {
override def sayHello(username: String): Unit = println(s"Hello, ${username}!")
override def shutdown(): Unit = println("Shutting down")
}
given Greeter = greeter
...
但我希望能够用一个表达式而不是两个表达式来做到这一点。相同的注释适用于先定义给定,然后将其绑定到名称:
given Greeter with {
override def sayHello(username: String): Unit = println(s"Hello, ${username}!")
override def shutdown(): Unit = println("Shutting down")
}
val greeter: Greeter = implicitly
特别是因为我认为implicitly
将在 3.1 中弃用并在 3.2 中消失。
我们还可以将给定对象的调用包装在函数中来解决这个问题:
def greet(username: String)(using g: Greeter): Unit = {
g.sayHello(username)
}
def shutdown(using greeter: Greeter): Unit = {
greeter.shutdown()
}
given Greeter with {
override def sayHello(username: String): Unit = println(s"Hello, ${username}!")
override def shutdown(): Unit = println("Shutting down")
}
try {
greet("world")
} finally {
shutdown
}
但这对我来说似乎是样板。