為什么 stride function 在 Swift 中會這樣工作？

Question

在第一個代碼片段下方打印 1

for i in stride(from: 1, through: 1, by: -1) {
    print(i)
}

但是為什么下面的代碼段什么也沒打印？

for i in stride(from: 1, through: 2, by: -1) {
    print(i)
}

Answer 1

對於正的步幅量， stride(from:through:by:)返回滿足x的所有值的序列

start <= x and x <= end

可以通過將步幅添加零次或多次來從起始值到達。 對於負步幅，條件是相反的：

start >= x and x >= end

這種行為關於正向和負向步幅是對稱的：

for i in stride(from: 0, through: 0, by: 1) { print(i) } // prints 0
for i in stride(from: 0, through: -1, by: 1) { print(i) } // prints nothing

for i in stride(from: 0, through: 0, by: -1) { print(i) } // prints 0
for i in stride(from: 0, through: 1, by: -1) { print(i) } // prints nothing

該實現可以在Swift源代碼庫中的 Stride.swift 中找到：

  public mutating func next() -> Element? {
    let result = _current.value
    if _stride > 0 ? result >= _end : result <= _end {
      // Note the `>=` and `<=` operators above. When `result == _end`, the
      // following check is needed to prevent advancing `_current` past the
      // representable bounds of the `Strideable` type unnecessarily.
      //
      // If the `Strideable` type is a fixed-width integer, overflowed results
      // are represented using a sentinel value for `_current.index`, `Int.min`.
      if result == _end && !_didReturnEnd && _current.index != .min {
        _didReturnEnd = true
        return result
      }
      return nil
    }
    _current = Element._step(after: _current, from: _start, by: _stride)
    return result
  }

你的第一個例子

for i in stride(from: 1, through: 1, by: -1) { print(i) } 

打印1因為該值滿足1 >= 1和1 >= 1 。 你的第二個例子

for i in stride(from: 1, through: 2, by: -1) { print(i) }

什么都不打印，因為沒有值同時滿足條件1 >= x和x >= 2 。