Scalaz(9)- typeclass:checking instance abiding the laws

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  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。

 

 


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