combineLatest(_:_:_:_:)

Subscribes to three additional publishers and invokes a closure upon receiving output from any of the publishers.

Declaration

func combineLatest<P, Q, R, T>(_ publisher1: P, _ publisher2: Q, _ publisher3: R, _ transform: @escaping (Self.Output, P.Output, Q.Output, R.Output) -> T) -> Publishers.Map<Publishers.CombineLatest4<Self, P, Q, R>, T> where P : Publisher, Q : Publisher, R : Publisher, Self.Failure == P.Failure, P.Failure == Q.Failure, Q.Failure == R.Failure

Parameters

publisher1:
A second publisher to combine with the first publisher.
publisher2:
A third publisher to combine with the first publisher.
publisher3:
A fourth publisher to combine with the first publisher.
transform:
A closure that receives the most-recent value from each publisher and returns a new value to publish.

Return Value

A publisher that receives and combines elements from this publisher and three other publishers.

Discussion

Use combineLatest(_:_:_:_:) when you need to combine the current and 3 additional publishers and transform the values using a closure in which you specify the published elements, to publish a new element.

The combined publisher passes through any requests to all upstream publishers. However, it still obeys the demand-fulfilling rule of only sending the request amount downstream. If the demand isn’t unlimited, it drops values from upstream publishers. It implements this by using a buffer size of 1 for each upstream, and holds the most-recent value in each buffer.

All upstream publishers need to finish for this publisher to finish. If an upstream publisher never publishes a value, this publisher never finishes.

In the example below, as combineLatest(_:_:_:_:) receives the most-recent values published by four publishers, multiplies them together, and republishes the result:

let pub = PassthroughSubject<Int, Never>()
let pub2 = PassthroughSubject<Int, Never>()
let pub3 = PassthroughSubject<Int, Never>()
let pub4 = PassthroughSubject<Int, Never>()

cancellable = pub
    .combineLatest(pub2, pub3, pub4) { firstValue, secondValue, thirdValue, fourthValue in
        return firstValue * secondValue * thirdValue * fourthValue
    }
    .sink { print("Result: \($0).") }

pub.send(1)
pub.send(2)
pub2.send(2)
pub3.send(9)
pub4.send(1)

pub.send(3)
pub2.send(12)
pub.send(13)
pub3.send(19)

// Prints:
//  Result: 36.     // pub = 2,  pub2 = 2,   pub3 = 9,  pub4 = 1
//  Result: 54.     // pub = 3,  pub2 = 2,   pub3 = 9,  pub4 = 1
//  Result: 324.    // pub = 3,  pub2 = 12,  pub3 = 9,  pub4 = 1
//  Result: 1404.   // pub = 13, pub2 = 12,  pub3 = 9,  pub4 = 1
//  Result: 2964.   // pub = 13, pub2 = 12,  pub3 = 19, pub4 = 1