斯威夫特： zip() 如何处理两个不同大小的集合？

Question

The zip() function takes two sequences and returns a sequence of tuples following:

output[i] = (sequence1[i], sequence2[i])

However, the sequences can potentially be of different dimensions. My question is how does the Swift language deal with this?

The docs were utterly useless.

Seems to me, there are two possibilities (in Swift):

• Stop at end of the shortest
• Stop at end of longest, filling with default constructor or a predefined value for shorter's element type

Answer 1

Swift uses the first option, the resulting sequence will have a length equal to the shorter of the two inputs .

For example:

let a: [Int] = [1, 2, 3]
let b: [Int] = [4, 5, 6, 7]

let c: [(Int, Int)] = zip(a, b) // [(1, 4), (2, 5), (3, 6)]

Answer 2

Apple's zip(_:_:) documentation has since been updated to answer this question:

https://developer.apple.com/documentation/swift/1541125-zip

If the two sequences passed to zip(_:_:) are different lengths, the resulting sequence is the same length as the shorter sequence . In this example, the resulting array is the same length as words :

 let words = ["one", "two", "three", "four"] let naturalNumbers = 1...Int.max let zipped = Array(zip(words, naturalNumbers)) // zipped == [("one", 1), ("two", 2), ("three", 3), ("four", 4)]