相关疑难解决方法(0)

我正在遵循Scala 和方法上的模式匹配和功能组合教程.有这样一个例子:composeandThen

scala> def addUmm(x: String) = x + " umm"
scala> def addAhem(x: String) = x + " ahem"

val ummThenAhem = addAhem(_).compose(addUmm(_))

当我尝试使用它时,我收到一个错误:

<console>:7: error: missing parameter type for expanded function ((x$1) => addAhem(x$1).compose(((x$2) => addUmm(x$2))))
   val ummThenAhem = addAhem(_).compose(addUmm(_))
                             ^
<console>:7: error: missing parameter type for expanded function ((x$2) => addUmm(x$2))
   val ummThenAhem = addAhem(_).compose(addUmm(_))
                                               ^
<console>:7: error: type mismatch;
 found   : java.lang.String
 required: Int
     val ummThenAhem = addAhem(_).compose(addUmm(_))

但是,这有效:

val ummThenAhem = …

我有两个功能,我正在努力compose:

  private def convert(value: String)
  : Path = decode[Path](value) match {


  private def verify(parsed: Path)
  : Path = parsed.os match {

我尝试过如下:

verify compose convert _

编译器抱怨:

[error] Unapplied methods are only converted to functions when a function type is expected.
[error] You can make this conversion explicit by writing `verify _` or `verify(_)` instead of `verify`.
[error]     verify compose convert _
[error]     ^
[error] one error found

我想完成以下事项:

  def process(value: String)
  : Path =
    verify(convert(value))

我究竟做错了什么？