为什么 Java8 和 Scala2.12 lambda 缓存之间有区别?

Question

Java 代码

package lambda_cache_example_java;
interface Semigroup1<A> {
public A append(A a1, A a2);
}
interface Semigroup2<A> {
public A append(A a1, A a2);
public interface Foo{}
public class Bar{}
}
class Main {
static Semigroup1<Integer> intSemigroup1() {
return (a1, a2) -> a1 + a2;
}
static Semigroup2<Integer> intSemigroup2() {
return (a1, a2) -> a1 + a2;
}
public static void main(String[] args) {
Semigroup1<Integer> x1 = intSemigroup1();
Semigroup1<Integer> x2 = intSemigroup1();
System.out.println(x1);
System.out.println(x2);
System.out.println(x1 == x2); // same instance
Semigroup2<Integer> y1 = intSemigroup2();
Semigroup2<Integer> y2 = intSemigroup2();
System.out.println(y1);
System.out.println(y2);
System.out.println(y1 == y2); // same instance as well
}
}

斯卡拉代码(版本 2.12.0)

package lambda_cache_example_scala
trait Semigroup1[A] {
def append(a1: A, a2: A): A
}
trait Semigroup2[A] {
def append(a1: A, a2: A): A
trait Foo
}
object Main {
def intSemigroup1(): Semigroup1[Int] =
(a1, a2) => a1 + a2
def intSemigroup2(): Semigroup2[Int] =
(a1, a2) => a1 + a2
def main(args: Array[String]): Unit = {
val x1 = intSemigroup1()
val x2 = intSemigroup1()
println(x1)
println(x2)
println(x1 eq x2) // same instance
val y1 = intSemigroup2()
val y2 = intSemigroup2()
println(y1)
println(y2)
println(y1 eq y2) // not same
}
}

结果

$ sbt "runMain lambda_cache_example_java.Main" "runMain lambda_cache_example_scala.Main"
[info] Running lambda_cache_example_java.Main 
lambda_cache_example_java.Main$$Lambda$9/1908283686@44939bb7
lambda_cache_example_java.Main$$Lambda$9/1908283686@44939bb7
true
lambda_cache_example_java.Main$$Lambda$10/2119574930@7f206457
lambda_cache_example_java.Main$$Lambda$10/2119574930@7f206457
true
[success] Total time: 0 s, completed 2016/11/24 15:09:56
[info] Running lambda_cache_example_scala.Main 
lambda_cache_example_scala.Main$$$Lambda$11/2085010450@7b408c6e
lambda_cache_example_scala.Main$$$Lambda$11/2085010450@7b408c6e
true
lambda_cache_example_scala.Main$$anonfun$intSemigroup2$2@c5329e5
lambda_cache_example_scala.Main$$anonfun$intSemigroup2$2@752d3cd9
false
[success] Total time: 0 s, completed 2016/11/24 15:09:57

Answer 1

Scala 具有路径依赖类型。虽然从您的示例中不明显，但可以构建嵌套特征，其中特征Foo例如Semigroup2与另一个Semigroup2实例的Foo完全不兼容。这篇文章和这个答案似乎是对路径依赖类型的很好的解释。

这意味着Semigroup2的实例也由其内部特征定义，因此在引用其方法之一时必须进行闭包。由于每次我们尝试引用该方法时都会动态重新创建闭包，因此匿名函数不同也就不足为奇了。

在Java中，情况并非如此。Semigroup2<A>.Foo是一种类型(与 Scala 不同，在 Scala 中，您需要一个Semigroup[A]实例来标识类型Foo)。