从Scala代码使用Google番石榴时出现编译器错误 [英] compiler error when using Google guava from scala code

问题描述

我正在使用Scala代码中的Google Guava。当我尝试将Int用作示例中的键类型时，就会出现问题：

I m using Google Guava from a scala code. And an issue occurs when I m trying to use Int as a key type like in the example:

CacheBuilder.newBuilder()
    .maximumSize(2)
    .expireAfterWrite(24, TimeUnit.HOURS)
    .build(
      new CacheLoader[Int, String] {
        def load(path: Int): String = {
          path + "hello"
        }
      }
    )

这似乎很好，但是所创建对象的推断类型为 LoadingCache [Int with AnyRef，String] ：

It seems to be fine, but the inferred type of created object is LoadingCache[Int with AnyRef, String]:

  val cache: LoadingCache[Int with AnyRef, String] = CacheBuilder.newBuilder()
        .maximumSize(2)
        .expireAfterWrite(24, TimeUnit.HOURS)
        .build(
          new CacheLoader[Int, String] {
            def load(path: Int): String = {
              path + "hello"
            }
          }
        )

当我尝试获取错误时会发生错误这样的元素：

And the error occurs when I m trying to get an element like in this example:

cache.get(1)

Scala编译器错误：

Scala compiler error:

[ERROR] error: type mismatch;
[INFO]  found   : Int(1)
[INFO]  required: Int
[INFO]   cache.get(1)
[INFO]             ^

有人可以指出为什么出现这种错误以及我在做什么错吗？

Can someone point me out why such error appears and what I m doing wrong?

ENV：

• Google Guava 15.0


• Scala 2.11.5

推荐答案

在 1 上不是 Int with AnyRef

问题中的编译错误与番石榴无关。此代码段在此处产生相同的错误：

On 1 not being an Int with AnyRef

The compile error in your question doesn't have anything to do with Guava. This snippet here produces the same error:

val h = new scala.collection.mutable.HashMap[Int with AnyRef, String]
h(3) = "hello"
println("Get 3: " + h.get(3))

给予

error: type mismatch;
found   : Int(3)
required: Int

这是由<$ c引起的$ c> Int with AnyRef ：因为 Int 是 AnyVal 的子类型，所以交集 Int with AnyRef 为空，根本就不存在该类型的实例。

This is caused by the Int with AnyRef: since Int is subtype of AnyVal, the intersection Int with AnyRef is empty, there simply cannot exist any instances with that type.

问题是，当您调用 .build（）时，scala编译器无法找到将用作 .build [Int，String] ，因为没有拆箱整数的版本。因此，相反，编译器推断 .build [带有Int，String的AnyRef] ，并构建了无法使用的缓存结构。

The problem is that when you call .build(), the scala compiler cannot find a version that would work as .build[Int, String], because there is no version for unboxed integers. So instead, the compiler infers .build[AnyRef with Int, String], and builds an unusable cache structure.

为避免这种情况，请使用 java.lang.Integer 代替 Int 。这是在guava 15.0 scala 2.11上编译并运行的：

To avoid this, use java.lang.Integer instead of Int. This here compiles and runs with guava 15.0 scala 2.11:

import com.google.common.cache._
import java.util.concurrent._

val cache: LoadingCache[java.lang.Integer, String] = CacheBuilder.newBuilder()
  .maximumSize(2)
  .expireAfterWrite(24, TimeUnit.HOURS)
  .build[java.lang.Integer, String](
    new CacheLoader[java.lang.Integer, String] {
      def load(path: java.lang.Integer): String = {
        path + "hello"
      }
    }
  )

cache.put(42, "hello, world")
println(cache.get(42))

它应该与Scala的 Int无缝运行，因为scala会将 Int s自动装箱到 java.lang.Integer 中。

It should work seamlessly with scala's Int, because scala autoboxes Ints into java.lang.Integer anyway.

类似错误的答案：

1. Scala Guava类型不匹配问题

1. Scala Guava type mismatch issue

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