How do I write the following in Swift3?
for (f = first; f <= last; f += interval)
{
n += 1
}
This is my own attempt
for _ in 0.stride(to: last, by: interval)
{
n += 1
}
问问题
48363 次
5 回答
69
Swift 2.2 -> 3.0: Strideable:sstride(...)被全局stride(...)函数取代
在 Swift 2.2 中，我们可以（正如您自己尝试过的那样）使用蓝图（和默认实现）函数stride(through:by:)和stride(to:by:) 协议Strideable
/* Swift 2.2: stride example usage */ let from = 0 let to = 10 let through = 10 let by = 1 for _ in from.stride(through, by: by) { } // from ... through (steps: 'by') for _ in from.stride(to, by: by) { } // from ..< to (steps: 'by')
而在 Swift 3.0 中，这两个函数已被删除Strideable，取而代之的是全局函数stride(from:through:by:)和stride(from:to:by:); 因此上述等效的 Swift 3.0 版本如下
/* Swift 3.0: stride example usage */ let from = 0 let to = 10 let through = 10 let by = 1 for _ in stride(from: from, through: through, by: by) { } for _ in stride(from: from, to: to, by: by) { }
在您的示例中，您希望使用闭区间步幅替代stride(from:through:by:)，因为循环中的不变量for使用与小于或等于( <=) 的比较。IE
/* example values of your parameters 'first', 'last' and 'interval' */
let first = 0
let last = 10
let interval = 2
var n = 0
for f in stride(from: first, through: last, by: interval) {
print(f)
n += 1
} // 0 2 4 6 8 10
print(n) // 6
当然，我们仅将您的循环用作从循环到for的通道的示例，您可以自然地，对于您的具体示例，只需计算而不需要循环（）。forstridenn=1+(last-first)/interval
strideSwift 3.0：更复杂的迭代增量逻辑的替代方案
随着进化提案 SE-0094 的实施，Swift 3.0 引入了全局sequence函数：
stride对于具有更复杂的迭代增量关系的情况（本例中不是这种情况），这可能是一个合适的替代方案。
声明
func sequence<T>(first: T, next: @escaping (T) -> T?) -> UnfoldSequence<T, (T?, Bool)> func sequence<T, State>(state: State, next: @escaping (inout State) -> T?) -> UnfoldSequence<T, State>
我们将简要介绍这两个函数中的第一个。参数采用一个闭包，该next闭包应用一些逻辑来懒惰地构造给定当前元素的下一个序列元素（以 开头first）。next序列在返回时终止nil，如果 anext永不返回，则序列终止nil。
应用于上面简单的恒定步幅示例，该sequence方法有点冗长和过度使用适合此目的的stride解决方案：
let first = 0
let last = 10
let interval = 2
var n = 0
for f in sequence(first: first,
next: { $0 + interval <= last ? $0 + interval : nil }) {
print(f)
n += 1
} // 0 2 4 6 8 10
print(n) // 6
然而，这些sequence函数对于具有非常量步幅的情况非常有用，例如在以下问答中涵盖的示例中：
只需注意以最终nil返回（如果不是：“无限”元素生成）来终止序列，或者，当 Swift 3.1 到来时，将其惰性生成与prefix(while:)序列方法结合使用，如进化提案SE-中所述0045 . 后者应用于此答案的运行示例使该sequence方法不那么冗长，显然包括元素生成的终止标准。
/* for Swift 3.1 */
// ... as above
for f in sequence(first: first, next: { $0 + interval })
.prefix(while: { $0 <= last }) {
print(f)
n += 1
} // 0 2 4 6 8 10
print(n) // 6
于 2016-05-16T09:20:29.783 回答
37
使用 Swift 5，您可以选择以下 5 个示例之一来解决您的问题。
#1。使用stride(from:to:by:)功能
let first = 0
let last = 10
let interval = 2
let sequence = stride(from: first, to: last, by: interval)
for element in sequence {
print(element)
}
/*
prints:
0
2
4
6
8
*/
#2。使用sequence(first:next:)功能
let first = 0
let last = 10
let interval = 2
let unfoldSequence = sequence(first: first, next: {
$0 + interval < last ? $0 + interval : nil
})
for element in unfoldSequence {
print(element)
}
/*
prints:
0
2
4
6
8
*/
#3。使用AnySequence init(_:)初始化器
let anySequence = AnySequence<Int>({ () -> AnyIterator<Int> in
let first = 0
let last = 10
let interval = 2
var value = first
return AnyIterator<Int> {
defer { value += interval }
return value < last ? value : nil
}
})
for element in anySequence {
print(element)
}
/*
prints:
0
2
4
6
8
*/
#4。使用CountableRange filter(_:)方法
let first = 0
let last = 10
let interval = 2
let range = first ..< last
let lazyCollection = range.lazy.filter({ $0 % interval == 0 })
for element in lazyCollection {
print(element)
}
/*
prints:
0
2
4
6
8
*/
#5。使用CountableRange flatMap(_:)方法
let first = 0
let last = 10
let interval = 2
let range = first ..< last
let lazyCollection = range.lazy.compactMap({ $0 % interval == 0 ? $0 : nil })
for element in lazyCollection {
print(element)
}
/*
prints:
0
2
4
6
8
*/
于 2017-06-29T11:31:14.883 回答
15
简而言之，Swift 3.0 的工作代码：
let (first, last, interval) = (0, 100, 1)
var n = 0
for _ in stride(from: first, to: last, by: interval) {
n += 1
}
于 2017-03-15T19:45:55.073 回答
0
我们也可以使用while循环作为替代方式
while first <= last {
first += interval
}
于 2020-04-04T16:43:27.267 回答
-5
for _ in 0.stride(to: last, by: interval) { n += 1 }
于 2016-05-11T18:18:45.233 回答