Question

在C＃中有一个非常方便的东西叫做方法组，基本上不是写：

someset.Select((x,y) => DoSomething(x,y))

你可以写：

someset.Select(DoSomething)

Scala中有类似内容吗？

例如：

int DoSomething(int x, int y)
{
    return x + y;
}

int SomethingElse(int x, Func<int,int,int> f)
{
    return x + f(1,2);
}

void Main()
{
    Console.WriteLine(SomethingElse(5, DoSomething));
}

Answer 1

在scala中我们称之为函数;-)。 (x,y) => DoSomething(x,y)是一个匿名函数或闭包，但您可以传递与您调用的方法/函数的签名匹配的任何函数，在本例中为map。因此，例如在scala中，您可以简单地编写

List(1,2,3,4).foreach(println)

或

case class Foo(x: Int)
List(1,2,3,4).map(Foo) // here Foo.apply(_) will be called

Answer 2

经过一些实验后，我得出的结论是它在Scala中的作用与在C＃中的作用相同（不确定它是否实际上是相同的......）

这就是我想要实现的目标（玩Play！所以Scala对我来说是新手，不知道为什么这在我的视图中不起作用，但是当我在解释器中尝试时它工作正常）

def DoStuff(a: Int, b : Int) = a + b

def SomethingElse(x: Int, f (a : Int, b: Int) => Int)) = f(1,2) + x

SomethingElse(5, DoStuff)    
res1: Int = 8

Answer 3

您实际上可以使用部分函数模拟方法组的行为。但是，它可能不是推荐的方法，因为您强制在运行时发生任何类型错误，并且需要花费一些成本来确定要调用的过载。但是，这段代码是否符合您的要求？

object MethodGroup extends App {
   //The return type of "String" was chosen here for illustration
   //purposes only. Could be any type.
   val DoSomething: Any => String = {
        case () => "Do something was called with no args"
        case (x: Int) => "Do something was called with " + x
        case (x: Int, y: Int) => "Do something was called with " + (x, y)
    }

    //Prints "Do something was called with no args"
    println(DoSomething())

    //Prints "Do something was called with 10"
    println(DoSomething(10))

    //Prints "Do something was called with (10, -7)"
    println(DoSomething(10,-7))

    val x = Set((), 13, (20, 9232))
    //Prints the following... (may be in a different order for you)
    //Do something was called with no args
    //Do something was called with 13
    //Do something was called with (20, 9232)
    x.map(DoSomething).foreach(println)
}

Scala中C＃方法组的等价物是什么？

3 个答案: