merge(with:_:_:_:)
Combines elements from this publisher with those from four other publishers, delivering an interleaved sequence of elements.
Declaration
func merge<B, C, D, E>(with b: B, _ c: C, _ d: D, _ e: E) -> Publishers.Merge5<Self, B, C, D, E> where B : Publisher, C : Publisher, D : Publisher, E : Publisher, Self.Failure == B.Failure, Self.Output == B.Output, B.Failure == C.Failure, B.Output == C.Output, C.Failure == D.Failure, C.Output == D.Output, D.Failure == E.Failure, D.Output == E.OutputParameters
- b:
A second publisher.
- c:
A third publisher.
- d:
A fourth publisher.
- e:
A fifth publisher.
Return Value
A publisher that emits an event when any upstream publisher emits an event.
Discussion
Use merge(with:_:_:_:) when you want to receive a new element whenever any of the upstream publishers emits an element. To receive tuples of the most-recent value from all the upstream publishers whenever any of them emit a value, use combineLatest(_:_:_:). To combine elements from multiple upstream publishers, use zip(_:_:_:).
In this example, as merge(with:_:_:_:) receives input from the upstream publishers, it republishes the interleaved elements to the downstream:
let pubA = PassthroughSubject<Int, Never>()
let pubB = PassthroughSubject<Int, Never>()
let pubC = PassthroughSubject<Int, Never>()
let pubD = PassthroughSubject<Int, Never>()
let pubE = PassthroughSubject<Int, Never>()
cancellable = pubA
.merge(with: pubB, pubC, pubD, pubE)
.sink { print("\($0)", terminator: " " ) }
pubA.send(1)
pubB.send(40)
pubC.send(90)
pubD.send(-1)
pubE.send(33)
pubA.send(2)
pubB.send(50)
pubC.send(100)
pubD.send(-2)
pubE.send(33)
// Prints: "1 40 90 -1 33 2 50 100 -2 33"The merged publisher continues to emit elements until all upstream publishers finish. If an upstream publisher produces an error, the merged publisher fails with that error.