scala - 如何对自定义类型列表使用 sum？

Question

引用下面的代码片段:

class Foo(val b:BigDecimal) {
  def +(f:Foo) = new Foo(b+f.b) 
}
val l = List(new Foo(1), new Foo(2))

l.sum // gives error: could not find implicit value for parameter num: Numeric[Foo]

让它发挥作用的最简单方法是什么？我知道我必须定义一些隐式转换。

注意:类 Foo 位于库内，因此我无法编辑该代码。我需要在 Foo 之外执行此操作。

尽管有几个使用Numeric类型的示例，但我找不到这个特殊情况的答案

Answer 1

您可以定义必要的隐式:

  class Foo(val b:BigDecimal) {
    def +(f:Foo) = new Foo(b+f.b)
  }
  object Foo {
    implicit val numericFoo: Numeric[Foo] = new Numeric[Foo] {
      override def plus(x: Foo, y: Foo): Foo = new Foo(x.b + y.b)
      override def minus(x: Foo, y: Foo): Foo = new Foo(x.b - y.b)
      override def times(x: Foo, y: Foo): Foo = new Foo(x.b * y.b)
      override def negate(x: Foo): Foo = new Foo(-x.b)
      override def fromInt(x: Int): Foo = new Foo(x)
      override def toInt(x: Foo): Int = x.b.toInt
      override def toLong(x: Foo): Long = x.b.toLong
      override def toFloat(x: Foo): Float = x.b.toFloat
      override def toDouble(x: Foo): Double = x.b.toDouble
      override def compare(x: Foo, y: Foo): Int = x.b.compare(y.b)
    }
  }

或者只是

implicit val numericFoo: Numeric[Foo] = new Numeric[Foo] {
  override def plus(x: Foo, y: Foo): Foo = x + y
  override def minus(x: Foo, y: Foo): Foo = ???
  override def times(x: Foo, y: Foo): Foo = ???
  override def negate(x: Foo): Foo = ???
  override def fromInt(x: Int): Foo = new Foo(x)
  override def toInt(x: Foo): Int = ???
  override def toLong(x: Foo): Long = ???
  override def toFloat(x: Foo): Float = ???
  override def toDouble(x: Foo): Double = ???
  override def compare(x: Foo, y: Foo): Int = ???
}

或者您可以定义自己的扩展方法

  implicit class FooList(foos: List[Foo]) {
    def sum1: Foo = foos.foldLeft(new Foo(0))(_ + _)
  }

  l.sum1