Question

我如何在scala中编写isFunction函数，以便这样做：

def isFunction(x:Any) = /* SomeCode */

println(isFunction(isFunction _)) //true
println(isFunction("not a function")) //false

Answer 1

相当难看，但它确实有效：

def isFunction(x:Any) = x match {
  case _: Function0[_] => true
  case _: Function1[_, _] => true
  case _: Function2[_, _, _] => true
  ...
  case _: Function22[...] => true
  case _: PartialFunction[_, _] => true
  case _ => false
}

Answer 2

在scala中，您可以将函数视为具有公共apply方法的对象。我不熟悉新的scala 2.10反射API，但你总是可以使用传统的java方式：

def isFunction(x:Any) = x.getClass.getMethods.map(_.getName).exists{name => 
  name == "apply" || name.startsWith("apply$")
}

val set = Set(1, 2)
val str = "abc"
val func = { _:Int=> 1 }
val map = Map(1 -> 2)
val tuple = 1->2
val obj = new { def apply = 1 }
val obj2 = new { private def apply = 2 } 

assert(isFunction(set))
assert(!isFunction(str))
assert(isFunction(func))
assert(isFunction(map))
assert(!isFunction(tuple))
assert(isFunction(obj))
assert(!isFunction(obj2))

Answer 3

我不是您解决方案的特别粉丝，我同意Daniel Sobral的评论。如果我必须实现它，我会通过隐式转换

以类型安全的方式实现它

trait IsFunctionable {
  def isFunction : Boolean
}

object IsFunctionable {

  object IsFunction extends IsFunctionable {
    def isFunction = true
  }

  object IsNotFunction extends IsFunctionable {
    def isFunction = false
  }

  implicit def function0ToIsFunctionable[A<:Function0[_]](a:A):IsFunctionable = IsFunction
  implicit def function1ToIsFunctionable[A<:Function1[_,_]](a:A):IsFunctionable = IsFunction
  implicit def function2ToIsFunctionable[A<:Function2[_,_,_]](a:A):IsFunctionable = IsFunction
  // and so on

  implicit def anyToIsFunctionable[A](a:A):IsFunctionable = IsNotFunction
}

现在你可以愉快地在repl中测试：

scala>  import IsFunctionable._
import IsFunctionable._

scala>  val a: Int => Int = _ * 2
a: Int => Int = <function1>

scala>  val b: (Int,Int) => Int = (_*_)
b: (Int, Int) => Int = <function2>

scala>  a(3)
res0: Int = 6
scala>  b(2,4)
res3: Int = 8

scala>  a.isFunction
res4: Boolean = true

scala>  b.isFunction
res5: Boolean = true

scala>  "Hello".isFunction
res6: Boolean = false

如何在scala中编写函数isFunction？

3 个答案: