类型 Int 不符合协议序列

Question

I have the following code in Swift 3:

var numbers = [1,2,1]
for number in numbers.count - 1 { // error
    if numbers[number]  < numbers[number + 1] {
        print(number)
    }
}

I am checking if the value on the index [number] is always higher than the value on the index [number + 1]. I am getting an error:

Type Int does not conform to protocol sequence

Any idea?

Answer 1

It may be swift. You can use this iteration.

for number in 0..<(numbers.count-1)

Answer 2

The error is because Int is not a Sequence . You can create a range as already suggested, which does conform to a sequence and will allow iteration using for in .

One way to make Int conform to a sequence is:

extension Int: Sequence {
    public func makeIterator() -> CountableRange<Int>.Iterator {
        return (0..<self).makeIterator()
    }
}

Which would then allow using it as a sequence with for in .

for i in 5 {
    print(i)
}

but I wouldn't recommend doing this. It's only to demonstrate the power of protocols but would probably be confusing in an actual codebase.

From you example, it looks like you are trying to compare consecutive elements of the collection. A custom iterator can do just that while keeping the code fairly readable:

public struct ConsecutiveSequence<T: IteratorProtocol>: IteratorProtocol, Sequence {
    private var base: T
    private var index: Int
    private var previous: T.Element?

    init(_ base: T) {
        self.base = base
        self.index = 0
    }

    public typealias Element = (T.Element, T.Element)

    public mutating func next() -> Element? {
        guard let first = previous ?? base.next(), let second = base.next() else {
            return nil
        }

        previous = second

        return (first, second)
    }
}

extension Sequence {
    public func makeConsecutiveIterator() -> ConsecutiveSequence<Self.Iterator> {
        return ConsecutiveSequence(self.makeIterator())
    }
}

which can be used as:

for (x, y) in [1,2,3,4].makeConsecutiveIterator() {
    if (x < y) {
        print(x)
    }
}

In the above example, the iterator will go over the following pairs:

(1, 2)
(2, 3)
(3, 4)

Answer 3

This maybe a little late but you could have done:

for number in numbers { }

instead of:

for number in numbers.count - 1 { }

For a for loop to work a sequence (range) is needed. A sequence consists of a stating a value, an ending value and everything in between. This means that a for loop can be told to loop through a range with ether

for number in 0...numbers.count-1 { }   `or`   for number in numbers { } 

Both example give the nesasery sequences. Where as:

 for number in numbers.count - 1 { }

Only gives one value that could either be the starting or the ending value, making it impossible to work out how many time the for loop will have to run.

For more information see Apple's swift control flow documnetation

Answer 4

This error can also come about if you try to enumerate an array instead of the enumerated array. For example:

for (index, element) in [0, 3, 4] { 

}

Should be:

for (index, element) in [0, 3, 4].enumerated() { 

}

Answer 5

So first you need to understand what is sequence.. A type that provides sequential, iterated access to its elements.

A sequence is a list of values that you can step through one at a time. The most common way to iterate over the elements of a sequence is to use a for-in loop:

let oneTwoThree = 1...3. // Sequence

for loop actually means

For number in Sequences {}

So you need to use for number in 0..<(numbers.count-1) {}

Answer 6

The error is because number is not an index, but the element of the array on each iteration. You can modify your code like this:

var numbers = [1,2,1,0,3]
for number in 0..<numbers.count - 1 {
    if numbers[number] < numbers[number + 1] {
        print(numbers[number])
    }
}

Or there is a trick using the sort method, but that's kind of a hack (and yes, the subindexes are right, but look like inverted; you can try this directly on a Playground):

var numbers = [1,2,1,0,3]
numbers.sort {
    if $0.1 < $0.0 {
        print ($0.1)
    }
    return false
}

Answer 7

For me, this error occurred when I tried writing a for loop, not for an array but a single element of the array.

For example:

let array = [1,2,3,4]
let item = array[0]

for its in item
{
   print(its)
}

This gives an error like: Type Int does not conform to protocol 'sequence'

So, if you get this error in for loop, please check whether you are looping an array or not.