Global function sequence(state:next:) and type inf

2019-01-20 12:24发布

Background & details

Swift evolution proposal SE-0094 was implemented in Swift 3.0, introducing the global sequence functions:

The latter is declared as follows

func sequence<T, State>(state: State, 
                        next: @escaping (inout State) -> T?) ->
         UnfoldSequence<T, State>

and is implemented in swift/stdlib/public/core/UnfoldSequence.swift. The language reference gives the following example for using it (note the lack of explicit type annotation)

// Interleave two sequences that yield the same element type
sequence(state: (false, seq1.makeIterator(), seq2.makeIterator()), next: { iters in
  iters.0 = !iters.0
  return iters.0 ? iters.1.next() : iters.2.next()
})

I cannot, however, get the example above to work (e.g. using let seq1 = 1...3, let seq2 = 4...6), but is prompted with the rather curious error message

error: ambiguous reference to member 'sequence(first:next:)'

Only if I explicitly type annotate the mutable State parameter in the next closure, as well as the return type of it, does the example above compile

let seq1 = 1...3
let seq2 = 4...6

for i in sequence(state: (false, seq1.makeIterator(), seq2.makeIterator()), 
        next: { (iters: inout (Bool, ClosedRangeIterator<Int>, ClosedRangeIterator<Int>)) 
                        -> Int? in
    iters.0 = !iters.0
    return iters.0 ? iters.1.next() : iters.2.next()
}) {
    print(i)
} // 1 4 2 5 3 6

This is not the way I hope to use sequence(state:next:), however, as I'd rather see it in on-the-fly applications where type inference works as it should, avoiding all the explicitness.

Question

  • Are we intended to use the sequence(first:next:) function with explicit type annotations as above? Is there some limitation in this function due to the inout parameter closure, or am I missing something?

2条回答
混吃等死
2楼-- · 2019-01-20 12:52

This looks like a combination of two issues.

The first is that Swift currently doesn't infer the type of a multi-line closure without any external context. This is however intended behaviour, as confirmed by Apple developer Jordan Rose in the comments of SR-1570:

This is correct behavior: Swift does not infer parameter or return types from the bodies of multi-statement closures. But the diagnostic could be a lot better.

Therefore in theory, you would just need to explicitly define the return type of the closure you pass to sequence()'s next: parameter, as the parameter type can be inferred from external context (namely the type you pass into the state: parameter):

let seq1 = 1...3
let seq2 = 4...6

let combined = sequence(state: (false, seq1.makeIterator(), seq2.makeIterator()),
                        next: { iters -> Int? in
    iters.0 = !iters.0
    return iters.0 ? iters.1.next() : iters.2.next()
})

(Edit: This now compiles in Swift 3.1)


However, this still doesn't compile – which is due to the second issue, where the compiler cannot infer the type for an inout closure parameter in Swift 3 (which wasn't the case in Swift 2). This is a suspected bug, which has already been filed (see both SR-1976 & SR-1811).

Therefore, as you note in the question, this means (quite unsatisfactorily) that you have to explicitly annotate the full closure signature that you pass to next::

let combined = sequence(state: (false, seq1.makeIterator(), seq2.makeIterator()),
                        next: { (iters: inout (Bool, ClosedRangeIterator<Int>, ClosedRangeIterator<Int>)) -> Int? in
    iters.0 = !iters.0
    return iters.0 ? iters.1.next() : iters.2.next()
})
查看更多
3楼-- · 2019-01-20 13:11

But even admitting that this is a bug, you can still make the code look a lot nicer instead of worrying about it (typed in the browser without testing, but something like this should work):

typealias MyTriple = (Bool, ClosedRangeIterator<Int>, ClosedRangeIterator<Int>)
let someTriple : MyTriple = (false, seq1.makeIterator(), seq2.makeIterator())
let combined = sequence(state: someTriple) { 
    (iters: inout MyTriple) -> Int? in
    iters.0 = !iters.0
    return iters.0 ? iters.1.next() : iters.2.next()
}
查看更多
登录 后发表回答