为什么 Scala 'String' 对象的迭代器和 'List[Int]' 对象的迭代器在这里表现不同?
Why Scala 'String' object's iterator and 'List[Int]' object's iterator are behaving differently here?
我只是想探索 REPL 中 String
对象和 List[Int]
对象的迭代器的行为,测试如下所示:
scala> val list = List(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16)
list: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16)
scala> val itL = list.iterator
itL: Iterator[Int] = non-empty iterator
scala> List(8,5,list.size-13).map(itL.take(_).mkString)
res85: List[String] = List(12345678, 910111213, 141516)
scala> val st = "abcdefghijklmnop"
st: String = abcdefghijklmnop
scala> val itS = st.iterator
itS: Iterator[Char] = non-empty iterator
scala> List(8,5,st.size-13).map(itS.take(_).mkString)
res84: List[String] = List(abcdefgh, abcde, abc)
为什么迭代器的行为不同?我在 String
对象的情况下的预期输出是:
List[String] = List(abcdefgh, ijklm, nop)
如果可能的话,谁能用例子解释一下。
另一个观察是:Range
对象的迭代器的行为也与 String
对象完全相似
如下所示:
scala> val x = (1 to 16)
x: scala.collection.immutable.Range.Inclusive = Range 1 to 16
scala> val t = (1 to 16).iterator
t: Iterator[Int] = non-empty iterator
scala> List(8,5,x.size-13).map(t.take(_).mkString)
res103: List[String] = List(12345678, 12345, 123)
如果将 Range
转换为 List
或 Set
,则各个迭代器的行为始终完全符合我的预期:
scala> val x1 = (1 to 16).toList
x1: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16)
scala> val t1 = x1.iterator
t1: Iterator[Int] = non-empty iterator
scala> List(8,5,x1.size-13).map(t1.take(_).mkString)
res104: List[String] = List(12345678, 910111213, 141516)
scala> val x2 = (1 to 16).toSet
x2: scala.collection.immutable.Set[Int] = Set(5, 10, 14, 1, 6, 9, 13, 2, 12, 7, 3, 16, 11, 8, 4, 15)
scala> val t2 = x2.iterator
t2: Iterator[Int] = non-empty iterator
scala> List(8,5,x2.size-13).map(t2.take(_).mkString)
res105: List[String] = List(51014169132, 12731611, 8415)
Iterator.take(n:Int)
API documentation:
附有注释
Reuse: After calling this method, one should discard the iterator it was called on, and use only the iterator that was returned. Using the old iterator is undefined, subject to change, and may result in changes to the new iterator as well.
您似乎发现了一些 "undefined" 行为。
我只是想探索 REPL 中 String
对象和 List[Int]
对象的迭代器的行为,测试如下所示:
scala> val list = List(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16)
list: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16)
scala> val itL = list.iterator
itL: Iterator[Int] = non-empty iterator
scala> List(8,5,list.size-13).map(itL.take(_).mkString)
res85: List[String] = List(12345678, 910111213, 141516)
scala> val st = "abcdefghijklmnop"
st: String = abcdefghijklmnop
scala> val itS = st.iterator
itS: Iterator[Char] = non-empty iterator
scala> List(8,5,st.size-13).map(itS.take(_).mkString)
res84: List[String] = List(abcdefgh, abcde, abc)
为什么迭代器的行为不同?我在 String
对象的情况下的预期输出是:
List[String] = List(abcdefgh, ijklm, nop)
如果可能的话,谁能用例子解释一下。
另一个观察是:Range
对象的迭代器的行为也与 String
对象完全相似
如下所示:
scala> val x = (1 to 16)
x: scala.collection.immutable.Range.Inclusive = Range 1 to 16
scala> val t = (1 to 16).iterator
t: Iterator[Int] = non-empty iterator
scala> List(8,5,x.size-13).map(t.take(_).mkString)
res103: List[String] = List(12345678, 12345, 123)
如果将 Range
转换为 List
或 Set
,则各个迭代器的行为始终完全符合我的预期:
scala> val x1 = (1 to 16).toList
x1: List[Int] = List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16)
scala> val t1 = x1.iterator
t1: Iterator[Int] = non-empty iterator
scala> List(8,5,x1.size-13).map(t1.take(_).mkString)
res104: List[String] = List(12345678, 910111213, 141516)
scala> val x2 = (1 to 16).toSet
x2: scala.collection.immutable.Set[Int] = Set(5, 10, 14, 1, 6, 9, 13, 2, 12, 7, 3, 16, 11, 8, 4, 15)
scala> val t2 = x2.iterator
t2: Iterator[Int] = non-empty iterator
scala> List(8,5,x2.size-13).map(t2.take(_).mkString)
res105: List[String] = List(51014169132, 12731611, 8415)
Iterator.take(n:Int)
API documentation:
Reuse: After calling this method, one should discard the iterator it was called on, and use only the iterator that was returned. Using the old iterator is undefined, subject to change, and may result in changes to the new iterator as well.
您似乎发现了一些 "undefined" 行为。