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Arrays 带量程的快速开关_Arrays_Swift_Switch Statement_Tuples - Fatal编程技术网

Arrays 带量程的快速开关

Arrays 带量程的快速开关,arrays,swift,switch-statement,tuples,Arrays,Swift,Switch Statement,Tuples,我有一个Int的数组,希望找到每个实例所处的范围 let array: [Int] = [35,34,33,32,31,30,28,27,26,25,24,23,7,6,5,4,3,2,1] 我有另一个数组的元组的Int let tupleArray: [(Int, Int)] = [(35,30),(28,22), (21,15), (14,8), (7,0)] 我正在使用开关对它们进行迭代 for a in 0..<array.count { var iteration:

我有一个
Int
数组
,希望找到每个实例所处的范围

let array: [Int] = [35,34,33,32,31,30,28,27,26,25,24,23,7,6,5,4,3,2,1]
我有另一个
数组
元组
Int

let tupleArray: [(Int, Int)] = [(35,30),(28,22), (21,15), (14,8), (7,0)]
我正在使用
开关
对它们进行迭代

for a in 0..<array.count {
    var iteration: Int = 0
    switch array[a] {
    case tupleArray[iteration].0...tupleArray[iteration].1:
        print("Within Range")
    default:
        print("Next iteration")
        repeat {
             iteration++ 
        } while tupleArray[iteration].0 < array[a]
    } 
}

对于0中的一个..我很惊讶你居然可以运行代码。在我的Xcode 7.1中,它给出了一个运行时错误:

fatal error: Can't form Range with end < start
for a in array {
    if let range = (tupleArray.filter { $0.1 <= a && a <= $0.0 }).first {
        print("\(a) is within \(range)")
    } else {
        print("Cannot find a range for \(a)")
    }
}