Question

我在scala中创建了这个案例类和函数：

abstract class Building[T]
case class University[T](a: Building[T], b: Building[T], c: T) extends Building

def u[A,B](a: Building[A]): Building[B] = a match {
  case n: University[A] => University[B](n.a, n.b, n.c);
}

然而，它在编译时说明了这一点：

[错误] test.scala：357：类型不匹配; [错误]发现：test.abc.def.University [B] [error] required：test.abc.def.Building [B]

我做错了什么？

Answer 1

University必须延伸Building[T]，而不仅仅是Building

case class University[T](a: Building[T], b: Building[T], c: T) extends Building[T]

这是完整的代码

abstract class Building[T]
case class University[T](a: Building[T], b: Building[T], c: T) extends Building[T]

在u函数中，输出类型为Building[B]。您正在返回University[B]但University[B]需要Building[B] b，即Building[B]和c B。

n.a返回Building[A]而不是Building[B]，因此University[B]需要两个Building[B]和一个B。您无法通过n.a，n.b和n.c来创建university[B]。

为了使代码编译，我改变了输出类型并返回University[B]

def u[A,B](a: Building[A]): Building[A] = a match {
  case n: University[A] => University[A](n.a, n.b, n.c);
}

更正代码

abstract class Building[T]

case class University[T](a: Building[T], b: Building[T], c: T) extends Building[T]

def convert[A, B](a: A): B = ???

def bConvert[A, B](building: Building[A]): Building[B] = ???

def u[A,B](a: Building[A]): Building[B] = a match {
  case n: University[A] => University[B](bConvert(n.a), bConvert(n.b), convert(n.c));
}

Scala：案例类和模式匹配

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