问题
There are use cases where it is useful to create a copy of an object which is an instance of a case class of a set of case classes, which have a specific value in common.
For example let's consider the following case classes:
case class Foo(id: Option[Int])
case class Bar(arg0: String, id: Option[Int])
case class Baz(arg0: Int, id: Option[Int], arg2: String)
Then copy can be called on each of these case class instances:
val newId = Some(1)
Foo(None).copy(id = newId)
Bar("bar", None).copy(id = newId)
Baz(42, None, "baz").copy(id = newId)
As described here and here there is no simple way to abstract this like this:
type Copyable[T] = { def copy(id: Option[Int]): T }
// THIS DOES *NOT* WORK FOR CASE CLASSES
def withId[T <: Copyable[T]](obj: T, newId: Option[Int]): T =
obj.copy(id = newId)
So I created a scala macro, which does this job (almost):
import scala.reflect.macros.Context
object Entity {
import scala.language.experimental.macros
import scala.reflect.macros.Context
def withId[T](entity: T, id: Option[Int]): T = macro withIdImpl[T]
def withIdImpl[T: c.WeakTypeTag](c: Context)(entity: c.Expr[T], id: c.Expr[Option[Int]]): c.Expr[T] = {
import c.universe._
val currentType = entity.actualType
// reflection helpers
def equals(that: Name, name: String) = that.encoded == name || that.decoded == name
def hasName(name: String)(implicit method: MethodSymbol) = equals(method.name, name)
def hasReturnType(`type`: Type)(implicit method: MethodSymbol) = method.typeSignature match {
case MethodType(_, returnType) => `type` == returnType
}
def hasParameter(name: String, `type`: Type)(implicit method: MethodSymbol) = method.typeSignature match {
case MethodType(params, _) => params.exists { param =>
equals(param.name, name) && param.typeSignature == `type`
}
}
// finding method entity.copy(id: Option[Int])
currentType.members.find { symbol =>
symbol.isMethod && {
implicit val method = symbol.asMethod
hasName("copy") && hasReturnType(currentType) && hasParameter("id", typeOf[Option[Int]])
}
} match {
case Some(symbol) => {
val method = symbol.asMethod
val param = reify((
c.Expr[String](Literal(Constant("id"))).splice,
id.splice)).tree
c.Expr(
Apply(
Select(
reify(entity.splice).tree,
newTermName("copy")),
List( /*id.tree*/ )))
}
case None => c.abort(c.enclosingPosition, currentType + " needs method 'copy(..., id: Option[Int], ...): " + currentType + "'")
}
}
}
The last argument of Apply (see bottom of above code block) is a List of parameters (here: parameters of method 'copy'). How can the given id of type c.Expr[Option[Int]] be passed as named parameter to the copy method with the help of the new macro API?
In particular the following macro expression
c.Expr(
Apply(
Select(
reify(entity.splice).tree,
newTermName("copy")),
List(/*?id?*/)))
should result in
entity.copy(id = id)
so that the following holds
case class Test(s: String, id: Option[Int] = None)
// has to be compiled by its own
object Test extends App {
assert( Entity.withId(Test("scala rulz"), Some(1)) == Test("scala rulz", Some(1)))
}
The missing part is denoted by the placeholder /*?id?*/.
回答1:
Here's an implementation that's also a little more generic:
import scala.language.experimental.macros
object WithIdExample {
import scala.reflect.macros.Context
def withId[T, I](entity: T, id: I): T = macro withIdImpl[T, I]
def withIdImpl[T: c.WeakTypeTag, I: c.WeakTypeTag](c: Context)(
entity: c.Expr[T], id: c.Expr[I]
): c.Expr[T] = {
import c.universe._
val tree = reify(entity.splice).tree
val copy = entity.actualType.member(newTermName("copy"))
val params = copy match {
case s: MethodSymbol if (s.paramss.nonEmpty) => s.paramss.head
case _ => c.abort(c.enclosingPosition, "No eligible copy method!")
}
c.Expr[T](Apply(
Select(tree, copy),
params.map {
case p if p.name.decoded == "id" => reify(id.splice).tree
case p => Select(tree, p.name)
}
))
}
}
It'll work on any case class with a member named id, no matter what its type is:
scala> case class Bar(arg0: String, id: Option[Int])
defined class Bar
scala> case class Foo(x: Double, y: String, id: Int)
defined class Foo
scala> WithIdExample.withId(Bar("bar", None), Some(2))
res0: Bar = Bar(bar,Some(2))
scala> WithIdExample.withId(Foo(0.0, "foo", 1), 2)
res1: Foo = Foo(0.0,foo,2)
If the case class doesn't have an id member, withId will compile—it just won't do anything. If you want a compile error in that case, you can add an extra condition to the match on copy.
Edit: As Eugene Burmako just pointed out on Twitter, you can write this a little more naturally using AssignOrNamedArg at the end:
c.Expr[T](Apply(
Select(tree, copy),
AssignOrNamedArg(Ident("id"), reify(id.splice).tree) :: Nil
))
This version won't compile if the case class doesn't have an id member, but that's more likely to be the desired behavior anyway.
回答2:
This is the solution of Travis where all parts are put together:
import scala.language.experimental.macros
object WithIdExample {
import scala.reflect.macros.Context
def withId[T, I](entity: T, id: I): T = macro withIdImpl[T, I]
def withIdImpl[T: c.WeakTypeTag, I: c.WeakTypeTag](c: Context)(
entity: c.Expr[T], id: c.Expr[I]
): c.Expr[T] = {
import c.universe._
val tree = reify(entity.splice).tree
val copy = entity.actualType.member(newTermName("copy"))
copy match {
case s: MethodSymbol if (s.paramss.flatten.map(_.name).contains(
newTermName("id")
)) => c.Expr[T](
Apply(
Select(tree, copy),
AssignOrNamedArg(Ident("id"), reify(id.splice).tree) :: Nil))
case _ => c.abort(c.enclosingPosition, "No eligible copy method!")
}
}
}
来源：https://stackoverflow.com/questions/13446528/howto-model-named-parameters-in-method-invocations-with-scala-macros