Concat different Streams and print on Java

Question

I'm trying to print a concat Stream that includes String and Integer elements. I've try a few ways, but none of them works. I can't find a proper way to do it on internet. It's just a testing code as you can see:

import java.util.stream.*;
class testconcat{
    public static void main(String[] args){
        Stream<String> part1=Stream.of("Testing the ");
        Stream<String> part2=Stream.of("streams concat");
        Stream<String> part3=Stream.of(" on Java, dividing ");
        Stream<String> part4=Stream.of("this phrase in ");
        IntStream part5=IntStream.of(6);
        Stream<String> part6=Stream.of(" parts.");

        String phrase=Stream.concat(Stream.concat(Stream.concat(Stream.concat(Stream.concat(part1, part2), part3), part4), part5), part6);
        System.out.println(phrase);
    }
}

I know part5 is an Integer so I can't concat it on a regular way, so I also tried:

        IntStream part5=IntStream.of(6);
        Stream<String> part6=Stream.of(" parts.");

        String phrase=Stream.concat(IntStream.concat(Stream.concat(Stream.concat(Stream.concat(part1, part2), part3), part4), part5), part6);
        System.out.println(phrase);
    }
}

and also:

        Integer part5=6;
        String part6=(" parts.");

        String phrase=Stream.concat(Stream.concat(Stream.concat(Stream.concat(Stream.concat(part1, part2), part3), part4);
        System.out.println(phrase + Integer.toString(part5) + part6);
    }
}

None of which works. Is there a way to do that? Thanks in advance.

Answer 1

I'm trying to print a concat Stream that includes String and Integer elements

Wrong, IntStream is a specific stream of primitive int value(s), and it is not the same as Stream, and therefore, IntStream::boxed or IntStream::mapToObj must be invoked to produce Stream<?> which can be used in Stream::concat:

Stream<?> many = Stream
    .concat(part1, Stream
        .concat(part2, Stream
            .concat(part3, Stream
                .concat(part4, Stream
                    .concat(part5.boxed(), part6)
                )
            )
        )
    );

To get the result of the string concatenation as String, a terminal operation `Stream.collect(Collectors.joining())` should be applied after mapping all the objects in the streams into String using `String::valueOf` (or `Objects::toString`):
```java
String result = many.map(String::valueOf).collect(Collectors.joining());
System.out.println(result);
// -> Testing the streams concat on Java, dividing this phrase in 6 parts.

---

However, such verbose Stream concatenation is redundant, and the Streams can be joined using Stream.of + Stream::flatMap. In the example below, IntStream mapped to String immediately, so no extra Stream::map is needed:

Stream<String> many = Stream
    .of(part1, part2, part3, part4, part5.mapToObj(String::valueOf), part6)
    .flatMap(s -> s);

String res = many.collect(Collectors.joining());
System.out.println(res);
// -> Testing the streams concat on Java, dividing this phrase in 6 parts.

Answer 2

If I understood you correctly and you want to create a combined stream string, you can use something like:

Stream.concat(...):

Stream<Integer> integerStream = IntStream.range(72, 129).boxed(); // to convert to Stream<Integer>
Stream<String> stringStream =Stream.of(" String Stream");
String str = Stream.concat(integerStream.map(String::valueOf), stringStream).collect(Collectors.joining(","));
System.out.println(str);

Or

Stream.of(...)

Stream<Integer> integerStream = IntStream.range(72, 129).boxed(); // to convert to Stream<Integer>
Stream<String> stringStream =Stream.of(" String Stream");
String str = Stream.of(integerStream.map(String::valueOf), stringStream).flatMap(t -> t).collect(Collectors.joining(","));
        System.out.println(str);