不考虑Reset
方法和原始列表的可变性,NoveNext
和Current
的只读向前用法:
Initially, the enumerator is positioned before the first element in the collection. You must call the MoveNext
method to advance the enumerator to the first element of the collection before reading the value of Current
; otherwise, Current
is undefined.
Current
returns the same object until MoveNext
is called. MoveNext
sets Current
to the next element.
If MoveNext
passes the end of the collection, the enumerator is positioned after the last element in the collection and MoveNext
returns false. When the enumerator is at this position, subsequent calls to MoveNext
also return false. If the last call to MoveNext
returned false, Current
is undefined.
输入数据,并初始化枚举器:
let ls = [0;1;2] let enumerator = (Seq.ofList ls).GetEnumerator() let ls = []
此时调用Current
会抛出异常,印证第一段描述:
enumerator.Current //InvalidOperationException : 枚举尚未开始。请调用 MoveNext。
我们改正代码,连续运行,直到序列的末尾:
let ls = [0;1;2] let enumerator = (Seq.ofList ls).GetEnumerator() show <| enumerator.MoveNext() // true show enumerator.Current // 0 show <| enumerator.MoveNext() // true show enumerator.Current // 1 show <| enumerator.MoveNext() // true show enumerator.Current // 2
当再次调用MoveNext()
返回false,接上段代码:
show <| enumerator.MoveNext() // false
当MoveNext()
返回false之后,如果此时调用Current
会抛出异常:
show enumerator.Current // InvalidOperationException : 枚举已完成。
当MoveNext()
返回false之后,继续调用MoveNext()
不会抛出异常,它总是返回false:
show <| enumerator.MoveNext() // false show <| enumerator.MoveNext() // false show <| enumerator.MoveNext() // false show <| enumerator.MoveNext() // false
当输入序列为空时,行为如下:
let ls = [] let enumerator = (Seq.ofList ls).GetEnumerator() show <| enumerator.MoveNext() // false show enumerator.Current // InvalidOperationException : 枚举已完成。
javaScript的枚举器用法如下:
const iterable = ['a', 'b']; const iterator = iterable[Symbol.iterator](); iterator.next() //, { value: 'a', done: false } iterator.next() //, { value: 'b', done: false } iterator.next() //, { value: undefined, done: true }