concat

After completion of the original upstream the elements of the given source will be emitted.

Fan-in operators

Signature

Source.concatSource.concat Flow.concatFlow.concat

Description

After completion of the original upstream the elements of the given source will be emitted.

Both streams will be materialized together.

Note

The concat operator is for backwards compatibility reasons “detached” and will eagerly demand an element from both upstreams when the stream is materialized and will then have a one element buffer for each of the upstreams, this is most often not what you want, instead use concatLazy

Example

Scala
source
val sourceA = Source(List(1, 2, 3, 4))
val sourceB = Source(List(10, 20, 30, 40))

sourceA.concat(sourceB).runWith(Sink.foreach(println))
// prints 1, 2, 3, 4, 10, 20, 30, 40
Java
sourceimport org.apache.pekko.stream.javadsl.Keep;
import org.apache.pekko.stream.javadsl.Source;
import org.apache.pekko.stream.javadsl.Sink;

import java.util.*;

Source<Integer, NotUsed> sourceA = Source.from(Arrays.asList(1, 2, 3, 4));
Source<Integer, NotUsed> sourceB = Source.from(Arrays.asList(10, 20, 30, 40));
sourceA.concat(sourceB).runForeach(System.out::println, system);
// prints 1, 2, 3, 4, 10, 20, 30, 40

Reactive Streams semantics

emits when the current stream has an element available; if the current input completes, it tries the next one

backpressures when downstream backpressures

completes when all upstreams complete

1.1.2