Swift 3.0的替代方法是使用助手功能

public func sequence<T>(first: T, while condition: @escaping (T)-> Bool, next: @escaping (T) -> T) -> UnfoldSequence<T, T> {
    let nextState = { (state: inout T) -> T? in
        // Return `nil` if condition is no longer satisfied:
        guard condition(state) else { return nil }
        // Update current value _after_ returning from this call:
        defer { state = next(state) }
        // Return current value:
        return state
    }
    return sequence(state: first, next: nextState)
}

从Express for动态循环迅速变化：

for f in sequence(first: (0, 1), while: { $1 <= 50 }, next: { ($1, $0 + $1)}) {
    print(f.1)
}
// 1 1 2 3 5 8 13 21 34

注意，为了以包括所产生的序列中的零，只要更换的初始值(0, 1)由(1, 0)：

for f in sequence(first: (1, 0), while: { $1 <= 50 }, next: { ($1, $0 + $1)}) {
    print(f.1)
}
// 0 1 1 2 3 5 8 13 21 34

这使得“人为”检查

if pair.1 == 0 { pair.1 = 1; return 0 }

多余的。根本原因是斐波那契数可以归纳为负索引（https://en.wikipedia.org/wiki/Generalizations_of_Fibonacci_numbers）：

 ... -8, 5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 8, ...