这将以相反的顺序递减1。

let num1 = [1,2,3,4,5]
for item in nums1.enumerated().reversed() { 

    print(item.offset) // Print the index number: 4,3,2,1,0
    print(item.element) // Print the value :5,4,3,2,1

}

或者你可以用这个index value属性

let num1 = [1,2,3,4,5]
for (index,item) in nums1.enumerated().reversed() { 

    print(index) // Print the index number: 4,3,2,1,0
    print(item) // Print the value :5,4,3,2,1

}

对我来说，这是最好的办法。

var arrayOfNums = [1,4,5,68,9,10]

for i in 0..<arrayOfNums.count {
    print(arrayOfNums[arrayOfNums.count - i - 1])
}

Swift向前

for i in stride(from: 5, to: 0, by: -1) {
    print(i)
}
//prints 5, 4, 3, 2, 1

for i in stride(from: 5, through: 0, by: -1) {
    print(i)
}
//prints 5, 4, 3, 2, 1, 0

至于Swift 2.2, Xcode 7.3(2016年6月10日):

for (index,number) in (0...10).enumerate() {
    print("index \(index) , number \(number)")
}

for (index,number) in (0...10).reverse().enumerate() {
    print("index \(index) , number \(number)")
}

输出:

index 0 , number 0
index 1 , number 1
index 2 , number 2
index 3 , number 3
index 4 , number 4
index 5 , number 5
index 6 , number 6
index 7 , number 7
index 8 , number 8
index 9 , number 9
index 10 , number 10


index 0 , number 10
index 1 , number 9
index 2 , number 8
index 3 , number 7
index 4 , number 6
index 5 , number 5
index 6 , number 4
index 7 , number 3
index 8 , number 2
index 9 , number 1
index 10 , number 0

反转一个数组只需要一个步骤。reverse()

var arrOfnum = [1,2,3,4,5,6]
arrOfnum.reverse()

您可以使用reversed()方法轻松反转值。

var i:Int
for i in 1..10.reversed() {
    print(i)
}

reversed()方法反转值。