在编译时指定合法隐含和/或非法隐含? [英] Specifying Legal and/or Illegal Implicits at Compile-time?

问题描述

比方说，我有两个特定的object可以从中检索导入.假设两个对象都有多个我想使用的有用的导入.为了简化本示例，我仅包括1:

Let's say that I have two particular object's from which I retrieve imports. Assume that both objects have multiple, useful imports that I want to use. I'm only including 1 for simplicity of this example:

scala> object Implicits1 { implicit def good: String => Int = _ => 42 }
defined object Implicits1

scala> object Implicits2 { implicit def bad: String => Int = _ => 666 }
defined object Implicits2

然后，给定foo:

scala> def foo(x: Int): Int = x
foo: (x: Int)Int

我执行通配符导入以获取隐式:

I perform my wildcard imports to get implicits:

scala> import Implicits1._
import Implicits1._

scala> import Implicits2._
import Implicits2._

在REPL上运行foo(".")表示Implicits2.bad的隐式已解决:

Running foo(".") on the REPL shows that Implicits2.bad's implicit was resolved:

scala> foo(".")
res0: Int = 666

但是，我实际上想要的是Implicits1.good，而不是Implicits2.bad.

But, I actually wanted Implicits1.good, not Implicits2.bad.

som-snytt和Shadowlands向我介绍了如何处理通配符导入-通配符导入，然后隐藏特殊隐式?.

som-snytt and Shadowlands educated me on how to handle wildcard imports - Wildcard Import, then Hide Particular Implicit?.

但是，我可以在编译时指定允许或禁止特定的隐式吗?

However, can I specify at compile-time that particular implicits are allowed or forbidden?

推荐答案

在REPL中进行测试会产生错误的结果.如果两个导​​入都在一个文件中，则会得到一个模棱两可的隐式错误.除此之外

Testing this in REPL will give wrong results. If you had both imports in a file, you'd get an ambiguous implicit error. Beyond this

我可以在编译时指定允许特定的隐式

can I specify at compile-time that particular implicits are allowed

导入它们.

或禁止

不导入它们.如果

两个对象都有多个我想使用的有用导入

both objects have multiple, useful imports that I want to use

然后您可以说导入所有内容除了这些特定的标识符" :

import Implicits2.{bad => _, bad2 => _, _}

如果已将隐式对象引入作用域，则禁用该对象的唯一方法是通过阴影处理，如链接的问题中所述.

If you have brought an implicit into scope, the only way to disable it is by shadowing, as in the linked question.

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