相关疑难解决方法(0)

为什么类型推断失败？

scala> val xs = List(1, 2, 3, 3)
xs: List[Int] = List(1, 2, 3, 3)

scala> xs.toSet map(_*2)
<console>:9: error: missing parameter type for expanded function ((x$1) => x$1.$times(2))
       xs.toSet map(_*2)

但是,如果xs.toSet已分配,则编译.

scala> xs.toSet
res42: scala.collection.immutable.Set[Int] = Set(1, 2, 3)

scala> res42 map (_*2)
res43: scala.collection.immutable.Set[Int] = Set(2, 4, 6)

此外,走另一条路,转换为Set从List,并映射List规定.

scala> Set(5, 6, 7)
res44: scala.collection.immutable.Set[Int] = Set(5, 6, 7)

scala> res44.toList map(_*2)
res45: List[Int] = List(10, 12, 14)

这种情况有效:

Seq(fromDir, toDir) find (!_.isDirectory) foreach (println(_))

然而,这不是:

Seq(fromDir, toDir) find (!_.isDirectory) foreach (throw new Exception(_.toString))

编译以此错误结束:

error: missing parameter type for expanded function ((x$4) => x$4.toString)

现在,如果我这样写它再次编译:

Seq(fromDir, toDir) find (!_.isDirectory) foreach (s => throw new Exception(s.toString))

我相信有合理的解释;)