What trouble could bring assining reversed() to a swift array?

Question

I\'m wondering about the reversed() method on a swift Array:

var items = [\"a\", \"b\", \"c\"]

items = items.reversed()

the signature of the r

Answer 1

There are 3 overloads of reversed() for an Array in Swift 3:

1. Treating the Array as a RandomAccessCollection,
   func reversed() -> ReversedRandomAccessCollection<Self> (O(1))
2. Treating the Array as a BidirectionalCollection,
   func reversed() -> ReversedCollection<Self> (O(1))
3. Treating the Array as a Sequence,
   func reversed() -> [Self.Iterator.Element] (O(n))

By default, reversed() pick the RandomAccessCollection's overload and return a ReversedRandomAccessCollection. However, when you write

items = items.reversed()

you are forcing the RHS to return a type convertible to the LHS ([String]). Thus, only the 3rd overload that returns an array will be chosen.

That overload will copy the whole sequence (thus O(n)), so there is no problem overwriting the original array.

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Instead of items = items.reversed(), which creates a copy of the array, reverse that and copy it back, you could reach the same effect using the mutating function items.reverse(), which does the reversion in-place without copying the array twice.