说我有：
case class X[F, A](f: F => F, a: A)
然后建立我们：
case class X2[F,A,B](x1: X[F,A], x2: X[F,B]) {
type I = (A,B)
def i: I = (x1.a, x2.a)
def f = x2.f compose x1.f
}
case class X3[F,A,B,C](x1: X[F,A], x2: X[F,B], x3: X[F,C]) {
type I = (A,B,C)
def i: I = (x1.a, x2.a, x3.a)
def f = x3.f compose x2.f compose x1.f
}
case class X4[F,A,B,C,D](x1: X[F,A], x2: X[F,B], x3: X[F,C], x4: X[F,D]) {
type I = (A,B,C,D)
def i: I = (x1.a, x2.a, x3.a, x4.a)
def f = x4.f compose x3.f compose x2.f compose x1.f
}
case class X5[F,A,B,C,D,E] // and so on
Shapeless如何帮助避免上述重复，手动编码每个arity？
注意：结果中的类型成员I是解决方案的必要部分;它并不只是为了澄清i的返回类型。
此探索的下一步是扩展解决方案以支持类似
的内容case class Y[F,G,A,B]
case class Y2[F,G,A1,B1,A2,B2] (y1: Y[F,G,A1,B1], y2: Y[F,G,A2,B2])
case class Y3[F,G,A1,B1,A2,B2,A3,B3](y1: Y[F,G,A1,B1], y2: Y[F,G,A2,B2], y3: Y[F,G,A3,B3])