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# merge(with:_:_:)

Combines elements from this publisher with those from three other publishers, delivering an interleaved sequence of elements.

```
func merge<B, C, D>(with b: B, _ c: C, _ d: D) -> Publishers.Merge4<Self, B, C, D> where B : Publisher, C : Publisher, D : 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
```

## Parameters

`b`

A second publisher.

`c`

A third publisher.

`d`

A fourth publisher.

## Return Value

A publisher that emits an event when any upstream publisher emits an event.

## Discussion

Use [`merge(with:_:_:)`](/documentation/Combine/Publisher/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(_:_:_:)`](/documentation/Combine/Publisher/combineLatest(_:_:_:)-48buc).
To combine elements from multiple upstream publishers, use [`zip(_:_:_:)`](/documentation/Combine/Publisher/zip(_:_:_:)-16rcy).

In this example, as [`merge(with:_:_:)`](/documentation/Combine/Publisher/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>()

cancellable = pubA
    .merge(with: pubB, pubC, pubD)
    .sink { print("\($0)", terminator: " " )}

pubA.send(1)
pubB.send(40)
pubC.send(90)
pubD.send(-1)
pubA.send(2)
pubB.send(50)
pubC.send(100)
pubD.send(-2)

// Prints: "1 40 90 -1 2 50 100 -2 "
```

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.

---

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