方法采用 Seq[T] 返回 String 而不是 Seq[Char] [英] Method taking Seq[T] to return String rather than Seq[Char]

问题描述

我想实现接受任意 Seq[T] 并返回 Seq[T] 的方法.但是当提供 String 时，它也应该返回 String.

I would like to implement method that takes arbitrary Seq[T] and returns Seq[T] as well. But when String is provided it should also return String.

传递 String 由于从 String 到 WrappedString extends IndexedSeq[Char] 的一些隐式转换而起作用，但我得到 Seq[Char] 作为回报.是否有可能取回 String ?

Passing String works due to some implicit conversion from String to WrappedString extends IndexedSeq[Char], but I get Seq[Char] in return. Is it possible to get String back?

val sx: Seq[Int] = firstAndLast(List(1, 2, 3, 4))
val s1: Seq[Char] = firstAndLast("Foo Bar")
val s2: String = firstAndLast("Foo Bar")  //incompatible types error

def firstAndLast[T](seq: Seq[T]) = Seq(seq.head, seq.last)

firstAndLast() 实现无关紧要，只是一个例子.

firstAndLast() implementation is irrelevant, it is only an example.

推荐答案

是的，这是可能的.您必须要求其中一种花哨的 CanBuildFroms:

Yes, it is possible. You’ll have to require one of those fancy CanBuildFroms:

import scala.collection.generic.CanBuildFrom

def firstAndLast[CC, A, That](seq: CC)(implicit asSeq: CC => Seq[A], cbf: CanBuildFrom[CC, A, That]): That = {
  val b = cbf(seq)
  b.sizeHint(2)
  b += seq.head
  b += seq.last
  b.result
}

这也适用于数组.奖励:示例中的所有行都将按预期编译和工作.

This will also work with arrays. Bonus: all lines in your example will compile and work as expected.

这篇关于方法采用 Seq[T] 返回 String 而不是 Seq[Char]的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！