Scala 中参数化类型的声明中是否允许使用中缀符号？

Question

我一直在努力理解 this post 中所提问题的答案。

下面的代码再现了已接受答案中提出的解决方案。

object Finder {
  def find[T <: Node](name: String)(implicit e: T DefaultsTo Node): T = 
    doFind(name).asInstanceOf[T]
}    

sealed class DefaultsTo[A, B]
trait LowPriorityDefaultsTo {
  implicit def overrideDefault[A,B] = new DefaultsTo[A,B]
}
object DefaultsTo extends LowPriorityDefaultsTo {
   implicit def default[B] = new DefaultsTo[B, B]
}

如果隐式定义为

我会理解的

(implicit e:  DefaultsTo[T, Node])

但是它是用看起来像中缀表示法的东西定义的。我从来不知道类型 DefaultsTo[T, Node] 可以写成 T DefaultsTo Node。这是怎么回事？

Answer 1

But instead it is defined in something that looks like infix notation. I never knew that a type DefaultsTo[T,Node] could be written as "T DefaultsTo Node". Is this what is going on ?

是的。规则非常简单：类型必须恰好有 2 个参数。它主要用于具有 <:< 等符号名称的类型，但正如您所见，它也适用于字母数字名称。

Scala specification, 3.2.8 Infix Types:

InfixType     ::=  CompoundType {id [nl] CompoundType}
An infix type T1 op T2 consists of an infix operator op which gets applied to two type operands T1 and T2. The type is equivalent to the type application op[T1,T2]. The infix operator op may be an arbitrary identifier.

All type infix operators have the same precedence; parentheses have to be used for grouping. The associativity of a type operator is determined as for term operators: type operators ending in a colon ‘:’ are right-associative; all other operators are left-associative.

In a sequence of consecutive type infix operations t0 op t1 op2 … opn tn, all operators op1,…,opn must have the same associativity. If they are all left-associative, the sequence is interpreted as (…(t0 op1 t1) op2…) opn tn, otherwise it is interpreted as t0 op1 (t1 op2 (…opn tn)…).

Scala 中参数化类型的声明中是否允许使用中缀符号？

Is infix notation allowed in declarations of a parameterized type in Scala?

scala

implicit

infix-notation

parametric-polymorphism