trait Persisted {
def id: Long
}
如何实现一个接受任何样例类实例并返回混入该特质的副本的方法?
该方法的签名如下:
def toPersisted[T](instance: T, id: Long): T with Persisted
trait Persisted {
def id: Long
}
如何实现一个接受任何样例类实例并返回混入该特质的副本的方法?
该方法的签名如下:
def toPersisted[T](instance: T, id: Long): T with Persisted
这可以通过宏来实现(自 Scala 2.10.0-M3 起正式成为 Scala 的一部分)。以下是一个示例代码。
1) 我的宏生成一个本地类,该类继承提供的 case class 和 Persisted,很像 new T with Persisted
。然后它会缓存参数(以防止多次评估)并创建所创建类的实例。
2) 我是如何知道要生成什么树形结构的呢?我有一个简单的应用程序 parse.exe,它打印从输入代码解析后得到的 AST。因此,我只需调用 parse class Person$Persisted1(first: String, last: String) extends Person(first, last) with Persisted
,记录输出并在我的宏中重现它即可。parse.exe 是 scalac -Xprint:parser -Yshow-trees -Ystop-after:parser
的包装器。有不同的方法来探索 AST,请阅读 "Scala 2.10 中的元编程"。
3) 如果您在scalac作为参数中提供-Ymacro-debug-lite
,则可以对宏扩展进行合理性检查。在这种情况下,所有扩展都将被打印出来,您将能够更快地检测到代码生成错误。
编辑。更新了2.10.0-M7的示例
使用纯Scala无法实现您想要的功能。问题在于像以下这样的混入:
scala> class Foo
defined class Foo
scala> trait Bar
defined trait Bar
scala> val fooWithBar = new Foo with Bar
fooWithBar: Foo with Bar = $anon$1@10ef717
scala> fooWithBar.getClass
res3: java.lang.Class[_ <: Foo] = class $anon$1
有关更多信息,请参见Scala中的动态混入 - 是否可能?。
您可以找到一个最新的工作解决方案,该解决方案利用Scala 2.10.0-RC1的Toolboxes API作为SORM项目的一部分。
trait Persisted {
def key: String
}
实际的工作对象
import tools.nsc.interpreter.IMain
import tools.nsc._
import reflect.mirror._
object PersistedEnabler {
def toPersisted[T <: AnyRef](instance: T, key: String)
(implicit instanceTag: TypeTag[T]): T with Persisted = {
val args = {
val valuesMap = propertyValuesMap(instance)
key ::
methodParams(constructors(instanceTag.tpe).head.typeSignature)
.map(_.name.decoded.trim)
.map(valuesMap(_))
}
persistedClass(instanceTag)
.getConstructors.head
.newInstance(args.asInstanceOf[List[Object]]: _*)
.asInstanceOf[T with Persisted]
}
private val persistedClassCache =
collection.mutable.Map[TypeTag[_], Class[_]]()
private def persistedClass[T](tag: TypeTag[T]): Class[T with Persisted] = {
if (persistedClassCache.contains(tag))
persistedClassCache(tag).asInstanceOf[Class[T with Persisted]]
else {
val name = generateName()
val code = {
val sourceParams =
methodParams(constructors(tag.tpe).head.typeSignature)
val newParamsList = {
def paramDeclaration(s: Symbol): String =
s.name.decoded + ": " + s.typeSignature.toString
"val key: String" :: sourceParams.map(paramDeclaration) mkString ", "
}
val sourceParamsList =
sourceParams.map(_.name.decoded).mkString(", ")
val copyMethodParamsList =
sourceParams.map(s => s.name.decoded + ": " + s.typeSignature.toString + " = " + s.name.decoded).mkString(", ")
val copyInstantiationParamsList =
"key" :: sourceParams.map(_.name.decoded) mkString ", "
"""
class """ + name + """(""" + newParamsList + """)
extends """ + tag.sym.fullName + """(""" + sourceParamsList + """)
with """ + typeTag[Persisted].sym.fullName + """ {
override def copy(""" + copyMethodParamsList + """) =
new """ + name + """(""" + copyInstantiationParamsList + """)
}
"""
}
interpreter.compileString(code)
val c =
interpreter.classLoader.findClass(name)
.asInstanceOf[Class[T with Persisted]]
interpreter.reset()
persistedClassCache(tag) = c
c
}
}
private lazy val interpreter = {
val settings = new Settings()
settings.usejavacp.value = true
new IMain(settings, new NewLinePrintWriter(new ConsoleWriter, true))
}
private var generateNameCounter = 0l
private def generateName() = synchronized {
generateNameCounter += 1
"PersistedAnonymous" + generateNameCounter.toString
}
// REFLECTION HELPERS
private def propertyNames(t: Type) =
t.members.filter(m => !m.isMethod && m.isTerm).map(_.name.decoded.trim)
private def propertyValuesMap[T <: AnyRef](instance: T) = {
val t = typeOfInstance(instance)
propertyNames(t)
.map(n => n -> invoke(instance, t.member(newTermName(n)))())
.toMap
}
private type MethodType = {def params: List[Symbol]; def resultType: Type}
private def methodParams(t: Type): List[Symbol] =
t.asInstanceOf[MethodType].params
private def methodResultType(t: Type): Type =
t.asInstanceOf[MethodType].resultType
private def constructors(t: Type): Iterable[Symbol] =
t.members.filter(_.kind == "constructor")
private def fullyQualifiedName(s: Symbol): String = {
def symbolsTree(s: Symbol): List[Symbol] =
if (s.enclosingTopLevelClass != s)
s :: symbolsTree(s.enclosingTopLevelClass)
else if (s.enclosingPackageClass != s)
s :: symbolsTree(s.enclosingPackageClass)
else
Nil
symbolsTree(s)
.reverseMap(_.name.decoded)
.drop(1)
.mkString(".")
}
}
测试应用程序
import PersistedEnabler._
object Sandbox extends App {
case class Artist(name: String, genres: Set[Genre])
case class Genre(name: String)
val artist = Artist("Nirvana", Set(Genre("rock"), Genre("grunge")))
val persisted = toPersisted(artist, "some-key")
assert(persisted.isInstanceOf[Persisted])
assert(persisted.isInstanceOf[Artist])
assert(persisted.key == "some-key")
assert(persisted.name == "Nirvana")
assert(persisted == artist) // an interesting and useful effect
val copy = persisted.copy(name = "Puddle of Mudd")
assert(copy.isInstanceOf[Persisted])
assert(copy.isInstanceOf[Artist])
// the only problem: compiler thinks that `copy` does not implement `Persisted`, so to access `key` we have to specify it manually:
assert(copy.asInstanceOf[Artist with Persisted].key == "some-key")
assert(copy.name == "Puddle of Mudd")
assert(copy != persisted)
}
虽然在对象创建之后无法组合对象,但是您可以使用类型别名和定义结构体进行非常广泛的测试,以确定对象是否具有特定的组合:
type Persisted = { def id: Long }
class Person {
def id: Long = 5
def name = "dude"
}
def persist(obj: Persisted) = {
obj.id
}
persist(new Person)
def id:Long
的对象都可以被视为持久化对象。 object Persistable {
type Compatible = { def id: Long }
implicit def obj2persistable(obj: Compatible) = new Persistable(obj)
}
class Persistable(val obj: Persistable.Compatible) {
def persist() = println("Persisting: " + obj.id)
}
import Persistable.obj2persistable
new Person().persist()
def save[T](data: T): T with Persisted
,该方法依赖于问题描述中所述的方法。 - Nikita Volkov