在前几篇关于Functor和Applilcative typeclass的讨论中我们自定义了一个类型Configure，Configure类型的定义是这样的：
1 case class Configure[+A](get: A)
2 object Configure {
3 implicit val configFunctor = new Functor[Configure] {
4 def map[A,B](ca: Configure[A])(f: A => B): Configure[B] = Configure(f(ca.get))
5 }
6 implicit val configApplicative = new Applicative[Configure] {
7 def point[A](a: => A) = Configure(a)
8 def ap[A,B](ca: => Configure[A])(cfab: => Configure[A => B]): Configure[B] = cfab map {fab => fab(ca.get)}
9 }
10 }
通过定义了Configure类型的Functor和Applicative隐式实例（implicit instance），我们希望Configure类型既是一个Functor也是一个Applicative。那么怎么才能证明这个说法呢？我们只要证明Configure类型的实例能遵循它所代表的typeclass操作定律就行了。Scalaz为大部分typeclass提供了测试程序（scalacheck properties）。在scalaz/scalacheck-binding/src/main/scala/scalaz/scalacheck/scalazProperties.scala里我们可以发现有关functor scalacheck properties:
1 object functor {
2 def identity[F[_], X](implicit F: Functor[F], afx: Arbitrary[F[X]], ef: Equal[F[X]]) =
3 forAll(F.functorLaw.identity[X] _)
4
5 def composite[F[_], X, Y, Z](implicit F: Functor[F], af: Arbitrary[F[X]], axy: Arbitrary[(X => Y)],
6 ayz: Arbitrary[(Y => Z)], ef: Equal[F[Z]]) =
7 forAll(F.functorLaw.composite[X, Y, Z] _)
8
9 def laws[F[_]](implicit F: Functor[F], af: Arbitrary[F[Int]], axy: Arbitrary[(Int => Int)],
10 ef: Equal[F[Int]]) = new Properties("functor") {
11 include(invariantFunctor.laws[F])
12 property("identity") = identity[F, Int]
13 property("composite") = composite[F, Int, Int, Int]
14 }
15 }
可以看到：functor.laws[F[_]]主要测试了identity, composite及invariantFunctor的properties。在scalaz/Functor.scala文件中定义了这几条定律：
1 trait FunctorLaw extends InvariantFunctorLaw {
2 /** The identity function, lifted, is a no-op. */
3 def identity[A](fa: F[A])(implicit FA: Equal[F[A]]): Boolean = FA.equal(map(fa)(x => x), fa)
4
5 /**
6 * A series of maps may be freely rewritten as a single map on a
7 * composed function.
8 */
9 def composite[A, B, C](fa: F[A], f1: A => B, f2: B => C)(implicit FC: Equal[F[C]]): Boolean = FC.equal(map(map(fa)(f1))(f2), map(fa)(f2 compose f1))
10 }
11 。
我们在下面试着对那个Configure类型进行Functor实例和Applicative实例的测试：
1 import scalaz._
2 import Scalaz._
3 import shapeless._
4 import scalacheck.ScalazProperties._
5 import scalacheck.ScalazArbitrary._
6 import scalacheck.ScalaCheckBinding._
7 import org.scalacheck.{Gen, Arbitrary}
8 implicit def cofigEqual[A]: Equal[Configure[A]] = Equal.equalA
9 //> cofigEqual: [A#2921073]=> scalaz#31.Equal#41646[Exercises#29.ex1#59011.Confi
10 //| gure#2921067[A#2921073]]
11 implicit def configArbi[A](implicit a: Arbitrary[A]): Arbitrary[Configure[A]] =
12 a map { b => Configure(b) } //> configArbi: [A#2921076](implicit a#2921242: org#15.scalacheck#121951.Arbitra
13 //| ry#122597[A#2921076])org#15.scalacheck#121951.Arbitrary#122597[Exercises#29.
14 //| ex1#59011.Configure#2921067[A#2921076]]
除了需要的import外还必须定义Configure类型的Equal实例以及任意测试数据产生器（test data generator）configArbi[A]。我们先测试Functor属性：
1 functor.laws[Configure].check //>
2 + functor.invariantFunctor.identity: OK, passed 100 tests.
3 //|
4 + functor.invariantFunctor.composite: OK, passed 100 tests.
5 //|
6 + functor.identity: OK, passed 100 tests.
7 //|
8 + functor.composite: OK, passed 100 tests.
成功通过Functor定律测试。
再看看Applicative的scalacheck property：scalaz/scalacheck/scalazProperties.scala
1 object applicative {
2 def identity[F[_], X](implicit f: Applicative[F], afx: Arbitrary[F[X]], ef: Equal[F[X]]) =
3 forAll(f.applicativeLaw.identityAp[X] _)
4
5 def homomorphism[F[_], X, Y](implicit ap: Applicative[F], ax: Arbitrary[X], af: Arbitrary[X => Y], e: Equal[F[Y]]) =
6 forAll(ap.applicativeLaw.homomorphism[X, Y] _)
7
8 def interchange[F[_], X, Y](implicit ap: Applicative[F], ax: Arbitrary[X], afx: Arbitrary[F[X => Y]], e: Equal[F[Y]]) =
9 forAll(ap.applicativeLaw.interchange[X, Y] _)
10
11 def mapApConsistency[F[_], X, Y](implicit ap: Applicative[F], ax: Arbitrary[F[X]], afx: Arbitrary[X => Y], e: Equal[F[Y]]) =
12 forAll(ap.applicativeLaw.mapLikeDerived[X, Y] _)
13
14 def laws[F[_]](implicit F: Applicative[F], af: Arbitrary[F[Int]],
15 aff: Arbitrary[F[Int => Int]], e: Equal[F[Int]]) = new Properties("applicative") {
16 include(ScalazProperties.apply.laws[F])
17 property("identity") = applicative.identity[F, Int]
18 property("homomorphism") = applicative.homomorphism[F, Int, Int]
19 property("interchange") = applicative.interchange[F, Int, Int]
20 property("map consistent with ap") = applicative.mapApConsistency[F, Int, Int]
21 }
22 }
applicative.laws定义了4个测试Property再加上apply的测试property。这些定律（laws）在scalaz/Applicative.scala里定义了：
1 trait ApplicativeLaw extends ApplyLaw {
2 /** `point(identity)` is a no-op. */
3 def identityAp[A](fa: F[A])(implicit FA: Equal[F[A]]): Boolean =
4 FA.equal(ap(fa)(point((a: A) => a)), fa)
5
6 /** `point` distributes over function applications. */
7 def homomorphism[A, B](ab: A => B, a: A)(implicit FB: Equal[F[B]]): Boolean =
8 FB.equal(ap(point(a))(point(ab)), point(ab(a)))
9
10 /** `point` is a left and right identity, F-wise. */
11 def interchange[A, B](f: F[A => B], a: A)(implicit FB: Equal[F[B]]): Boolean =
12 FB.equal(ap(point(a))(f), ap(f)(point((f: A => B) => f(a))))
13
14 /** `map` is like the one derived from `point` and `ap`. */
15 def mapLikeDerived[A, B](f: A => B, fa: F[A])(implicit FB: Equal[F[B]]): Boolean =
16 FB.equal(map(fa)(f), ap(fa)(point(f)))
17 }
再测试一下Configure类型是否也遵循Applicative定律：
1 pplicative.laws[Configure].check //>
2 + applicative.apply.functor.invariantFunctor.identity: OK, passed 100 tests
3 //|
4 //| .
5 //|
6 + applicative.apply.functor.invariantFunctor.composite: OK, passed 100 test
7 //|
8 //| s.
9 //|
10 + applicative.apply.functor.identity: OK, passed 100 tests.
11 //|
12 + applicative.apply.functor.composite: OK, passed 100 tests.
13 //|
14 + applicative.apply.composition: OK, passed 100 tests.
15 //|
16 + applicative.identity: OK, passed 100 tests.
17 //|
18 + applicative.homomorphism: OK, passed 100 tests.
19 //|
20 + applicative.interchange: OK, passed 100 tests.
21 //|
22 + applicative.map consistent with ap: OK, passed 100 tests.
功通过了Applicative定律测试。现在我们可以说Configure类型既是Functor也是Applicative。
原创文章，作者：ItWorker，如若转载，请注明出处：https://blog.ytso.com/12938.html