Question

所以我有一个元组，我想作为Scala中的case类的参数传递。对于没有类型参数的case类，这很容易，我可以这样做：

 scala> case class Foo(a: Int, b: Int)
 defined class Foo

 scala> (Foo.apply _)
 res0: (Int, Int) => Foo = <function2>

 scala> val tuple = (1, 2)
 tuple: (Int, Int) = (1,2)

 scala> res0.tupled(tuple)
 res1: Foo = Foo(1,2)

 scala> Foo.tupled(tuple)
 res2: Foo = Foo(1,2)

但是，如果case类具有类型参数，则它似乎不起作用：

scala> (Bar.apply _)
res26: (Nothing, Nothing) => Bar[Nothing] = <function2>

scala> res26.tupled(tuple)
<console>:18: error: type mismatch;
 found   : (Int, Int)
 required: (Nothing, Nothing)
              res26.tupled(tuple)
                           ^
scala> (Bar[Int].apply _)
<console>:16: error: missing arguments for method apply in object Bar;
follow this method with `_' if you want to treat it as a partially applied function
              (Bar[Int].apply _)

scala> Bar.tupled(tuple)
<console>:17: error: value tupled is not a member of object Bar
              Bar.tupled(tuple)
                  ^

scala> Bar[Int].tupled(tuple)
<console>:17: error: missing arguments for method apply in object Bar;
follow this method with `_' if you want to treat it as a partially applied function
              Bar[Int].tupled(tuple)
                 ^

如何使用类型参数部分应用案例类？我正在使用Scala 2.10.3。

Answer 1

这似乎有效：

scala> case class Foo[X](a:X, b: X)
defined class Foo

scala> Foo.apply[Int] _
res1: (Int, Int) => Foo[Int] = <function2>

scala>

如何在Scala中部分应用带类型参数的case类

1 个答案: