从CaseClass(Int,String)创建函数[Int,String,CaseClass] #tupled？

Question

鉴于:

scala> case class Foo(x: Int, y: String)
defined class Foo

我正在尝试Foo.tupled用来创建Function2[Int, String, Foo]:

scala> val fn2: Function2[Int, String, Foo] = Foo.tupled match { 
     |   case (param1, param2)  => { (param1, param2) => Foo(param1, param2) } 
     | }
<console>:18: error: constructor cannot be instantiated to expected type;
 found   : (T1, T2)
 required: ((Int, String)) => Foo
         case (param1, param2)  => { (param1, param2) => Foo(param1, param2) }

但是,它不起作用.我该如何解决这个破碎的代码？

Answer 1

我不确定比赛应该做什么.你不需要它.要创建一个Function1[(Int, String), Foo],请使用如下的.tupled:

scala> case class Foo(x: Int, y: String)
defined class Foo

scala> val f = Foo.tupled
f: ((Int, String)) => Foo = <function1>

scala> f((1, "x"))
res0: Foo = Foo(1,x)

如果你想Function2[Int, String, Foo],你不需要使用.tupled 可言.具有N个参数的案例类的伴随对象已经实现了FunctionN特征.

scala> val x: Function2[Int, String, Foo] = Foo
x: (Int, String) => Foo = Foo

scala> :javap -c Foo$
Compiled from "<console>"
public class Foo$ extends scala.runtime.AbstractFunction2<java.lang.Object, java.lang.String, Foo> implements scala.Serializable {
  public static final Foo$ MODULE$;