像Scala中的Java 8中一样的方法引用 [英] Method References like in Java 8 in Scala

问题描述

在此Java类中:

import java.util.function.*;

public class T {

    public String func(String a) {
        System.out.println("There we go: " + a);
        return a;
        }

    public static void main(String... args) {
        final Supplier<T> c = T::new;
        final BiFunction<T, String, String> f = T::func;

        final T t = c.get();
        final String v = f.apply(t, "something");

        System.out.println(v);
    }

}

我可以获得对T的构造函数和实例方法func的方法引用.

I can get a method reference to the constructor of T and to the instance method func.

有没有办法在scala中做同样的事情，即获得

Is there a way to do the same in scala, i.e. to get

val c: () => T = ??? // default constructor of T as function
val f: (T, String) => String = ??? // instance method func of T as function

无需像这样包装它们:

val c: () => T = () => new T
val f: (T, String) => String = (t, arg) => t.func(arg)

即有没有一种方法像Java 8一样优雅，可以获取构造函数和实例方法引用来为这些事情获取scala函数?

i.e. is there a way which is as elegant as the Java 8 way to get constructor and instance method references to obtain scala functions for these things?

推荐答案

首先，让我们看一下Java代码到Scala的字面翻译:

First, let's have a look at a literal translation of the Java code to Scala:

class T {
  def func(a:String) : String = {
    println(s"There we go: $a")
    a
  }
}
object T {
  def main(args: Array[String]) = {
    val supplier = () => new T
    val f = (t:T) => t.func _

    val t = supplier()
    val v = f(t)("something")
    println(v)
  }
}

在Scala中，函数是一等公民，因此不需要像Java Supplier那样对生成的事物"进行特殊构造，因为它被建模为函数:f: () => T(对应的Consumer为f: T => ())

In Scala, functions are first class citizens, hence there's no need to have particular constructions for "things that generate", like the Java Supplier, as it's modeled as a function: f: () => T (same thing goes for its counterpart, the Consumer as f: T => ())

我们刚刚说过，函数是一等公民，所以让我们看一下使用该范式的上述版本:

We just said that functions are first class citizens, so let's see a version of the above using this paradigm:

object Tfunc {
  // let's remove the println side-effect from a function.
  val func: String => String = a => s"There we go: $a"

  def main(args: Array[String]) = {
    println(func("something"))
  }
}

在Scala中，没有对应的方法来获取构造函数引用，但是如果目标是使用功能方法，则Scala object提供了一种简单的构造来保存函数，而无需实例化.

In Scala, there's no counterpart to obtain a constructor reference, but if the aim is to use a functional approach, Scala objects offer a simple construct to hold functions and do not require to be instantiated.

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