Scala Quasikote宏示例断开类型签名关闭

我从书中选取了这个Scala Quasikote示例

我得到这个错误：

类型推断显示“Trees#Tree”，但类型推断已关闭

import scala.reflect.api.Trees // For Trees#Tree (TreeNode)
import scala.reflect.macros.blackbox._
import scala.reflect.runtime.universe._ // To use Scala runtime reflection

/**
  * Represents a macro invariant which is checked over the corresponding statements.
  * Example:
  * '''
  * var mustBeHello = "Hello"
  * invariant.execute(mustBeHello.equals("Hello")) {
  *   mustBeHello = "Goodbye"
  * }
  * // Throws invariant.InvariantFailure
  * '''
  */
object invariant {
  case class InvariantFailure(message: String) extends RuntimeException(message)

  type SyntaxTree = scala.reflect.runtime.universe.Tree

  type TreeNode = Trees#Tree // a syntax tree node that is in and of itself a tree

  // These two methods are the same, but one is a function call and the other is a macro function call
  def execute[RetType]              (myPredicate: => Boolean)(block: => RetType): RetType = macro executeMacro
  def executeMacro(context: Context)(myPredicate: SyntaxTree)(block: SyntaxTree) = {

    val predicateString: String = showCode(myPredicate) // turn this predicate into a String
    val q"..$statements" = block // make the block into a sequence of statements
    val myStatements: Seq[TreeNode] = statements // the statements are a sequence of SyntaxTreeNodes, each node a little Tree
    val invariantStatements = statements.flatMap { statement =>
        // Error here:
        val statementString: String = showCode(statement) /* Type mismatch, expected Tree, actual Trees#Tree */

        val message: String =
            s"FAILURE! $predicateString == false, for statement: " + statementString
        val tif: SyntaxTree =
            q"throw new metaprogramming.invariant.InvariantFailure($message)"
        val predicate2: SyntaxTree =
            q"if (false == $myPredicate) $tif"
        val toReturn: List[SyntaxTree] =
            List(q"{ val temp = $myStatements; $predicate2; temp };")
        toReturn
      }
    val tif: SyntaxTree =
        q"throw new metaprogramming.invariant.InvariantFailure($predicateString)"
    val predicate: SyntaxTree =
        q"if (false == $predicate) $tif"
    val toReturn: SyntaxTree =
        q"$predicate; ..$invariantStatements"
    toReturn
  }
}

^文件应该是不言自明的。类型推断为Tree#Tree，但在示例代码中添加“：Tree#Tree”会导致编译失败，并出现错误：

[info] Compiling 2 Scala sources to /home/johnreed/sbtProjects/scala-trace-debug/target/scala-2.11/test-classes...
[error] /home/johnreed/sbtProjects/scala-trace-debug/src/test/scala/mataprogramming/invariant2.scala:30: type mismatch;
[error] found : TreeNode
error scala.reflect.api.Trees#Tree
[error] required: context.universe.Tree
[error] val exceptionMessage = s"FAILURE! $predicateAsString == false, for statement: " + showCode(statement)

我要进IntelliJ了

[info] Compiling 2 Scala sources to /home/johnreed/sbtProjects/scala-trace-debug/target/scala-2.11/test-classes...
[error] /home/johnreed/sbtProjects/scala-trace-debug/src/test/scala/mataprogramming/invariant2.scala:30: type mismatch;
[error] found : TreeNode
error scala.reflect.api.Trees#Tree
[error] required: context.universe.Tree
[error] val exceptionMessage = s"FAILURE! $predicateAsString == false, for statement: " + showCode(statement)

这些类型真的很时髦。要么IntelliJ搞乱了类型，要么我不理解这个概念。

发现推断类型的“正常”方法是：

• 使用

  ：键入
  

• 分配到错误类型并观察错误消息

• 调用一个函数，该函数显示事物的

  类型标签

比如说,

[error] /home/apm/clones/prog-scala-2nd-ed-code-examples/src/main/scala/progscala2/metaprogramming/invariant2.scala:25: type mismatch;
[error]  found   : List[context.universe.Tree]
[error]  required: Int
[error]     val foo: Int = statements
[error]                    ^

这表明

树

在上下文中是路径依赖的

你不能用任何一棵老树来喂它

---

类似问题出现在。

代码来自何处？它不在回购协议中。这些类型是路径相关的，这就是为什么要使用

c.Tree

等。repo会编译。@som snytt-这段代码有些修改。来自repo的未修改代码将编译，但类型已关闭。看，你的代码不应该编译。当你指a.B.时，你不能使用类型项目a#B。我试试你的票证代码。