Scala TraversableOnce和toSet

scala> List(1).toSet()
res2: Boolean = false

scala> List(1).toSet.apply()
res3: Boolean = false

scalac具有称为“适应参数列表”的功能,这可以用-Xlint显示：

scala> List(1).toSet()
<console>:8: warning: Adapting argument list by inserting (): this is unlikely to be what you want.
        signature: GenSetLike.apply(elem: A): Boolean
  given arguments: <none>
 after adaptation: GenSetLike((): Unit)
              List(1).toSet()
                           ^
res7: Boolean = false

scalac尝试将参数包装成一个元组,就像sources中可以看到的那样(空的参数列表将被视为gen.mkTuple的单位文字)：

/* Try packing all arguments into a Tuple and apply `fun`
       * to that. This is the last thing which is tried (after
       * default arguments)
       */
      def tryTupleApply: Tree = (
        if (eligibleForTupleConversion(paramTypes,argslen) && !phase.erasedTypes) {
          val tupleArgs = List(atPos(tree.pos.makeTransparent)(gen.mkTuple(args)))
          // expected one argument,but got 0 or >1 ==> try applying to tuple
          // the inner "doTypedApply" does "extractUndetparams" => restore when it fails
          val savedUndetparams = context.undetparams
          silent(_.doTypedApply(tree,fun,tupleArgs,mode,pt)) map { t =>
              // Depending on user options,may warn or error here if
              // a Unit or tuple was inserted.
              val keepTree = (
                   !mode.typingExprNotFun
                || t.symbol == null
                || checkValidAdaptation(t,args)
              )
              if (keepTree) t else EmptyTree
          } orElse { _ => context.undetparams = savedUndetparams ; EmptyTree }
        }
        else EmptyTree
      )

哪个btw是not mentioned in the spec的一个功能.明确添加括号会让警告消失：

scala> List(1).toSet(())
res8: Boolean = false

现在还有一个问题是为什么上面的代码不会产生一个编译错误,因为列表是List [Int]的类型,而Set的apply方法具有类型签名apply(A)：Boolean,因此期望Int我们的情况.其原因是very well known problem和toSet [B>：A]的类型签名的结果：Set [B].类型签名表示下限,这意味着Int的任何超类型都可以作为参数传递.

因为在我们的例子中,Unit被指定为参数的类型,编译器必须搜索与toSet的类型签名相匹配的Unit和Int的公共超类型.并且因为有这样一个类型,即AnyVal,编译器会推断出这个类型,并且前进,而不会出现一个错误：

scala> List(1).toSet[AnyVal](())
res9: Boolean = false