scala - 如何声明一个scala case类为Scalaz的Semigroup的实例？

我有一个类型，定义如下:

import scalaz._, Scalaz._
case class MyInt(i : Int)

我想创建一个Semigroup的实例。我尝试了这个:

object MyInt {
  implicit def myIntSemigroup: Semigroup[MyInt] = new Semigroup[MyInt] {
    def append(a: MyInt, b: MyInt) : MyInt = MyInt(a.i + b.i)
  }
}

当我运行sbt console时，出现以下错误:

[info] Set current project to hello (in build file:/home/mp/code/scala/examples/semigroup/)
[info] Compiling 1 Scala source to /home/mp/code/scala/examples/semigroup/target/scala-2.10/classes...
[error] /home/mp/code/scala/examples/semigroup/src/main/scala/Main.scala:6: object creation impossible, since method append in trait Semigroup of type (f1: MyInt, f2: => MyInt)MyInt is not defined
[error]   implicit def myIntSemigroup: Semigroup[MyInt] = new Semigroup[MyInt] {
[error]                                                       ^
[error] one error found
[error] (compile:compile) Compilation failed
[error] Total time: 5 s, completed 25-Jun-2015 08:15:37

如何使MyInt成为Semigroup，以便可以在其上使用|+|？

为了完整起见，这是我的build.sbt文件:

name := "hello"
version := "1.0"
scalaVersion := "2.10.5"
libraryDependencies += "org.scalaz" %% "scalaz-core" % "7.1.3"

最佳答案

如果您仔细寻找Scalaz半组的签名，则第二个参数是按名称定义的，而在您的实现中您具有不同的签名。将您的第二个参数更改为b: => MyInt，它将编译