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I have written this lambda that is getting values from a map <Code, Message>contained inside an enum MessageKeyRegistry and creating adding to another map with Code as the key and the MessageKeyRegistry as the value.

Map<MessageKey, MessageKeyRegistry> keyMap = new HashMap<>();
    EnumSet.allOf(MessageKeyRegistry.class)
            .forEach(messageKeyRegistry -> messageKeyRegistry.getCodeToMessage().keySet()
                    .forEach(code -> keyMap.put(code, messageKeyRegistry)));

Now I want to satisfy the immutability concept of functional programming and write a lambda that returns an immutable map directly.

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  • please show the MessageKeyRegistry structure. Commented Apr 10, 2020 at 19:58

3 Answers 3

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I have written this lambda that is..

You didn't write just a lambda, but a chain of methods using lambda expressions. one if them is code -> keyMap.put(code, messageKeyRegistry).

Back to your question. You start with the iteration correctly (from the EnumSet), the next step is to realize what happens in the forEach methods.

  • Extract codes from the set of keys using them as keys using messageKeyRegistry
  • Using the extractor as a value messageKeyRegistry.

This is a job for map, however since the data structure is more complicated (collection in a collection), then flatMap would serve better. Finally, you'd result with something like:

Map<MessageKey, MessageKeyRegistry> keyMap = EnumSet.allOf(MessageKeyRegistry.class)
        .stream()
        .flatMap(messageKeyRegistry -> messageKeyRegistry.getCodeToMessage().keySet()
             .stream()
             .map(code -> new AbstractMap.SimpleEntry<>(code, messageKeyRegistry)))
        .collect(Collectors.toMap(Entry::getKey, Entry::getValue));

However, what you do has a drawback. I see the keys are overridden since there cannot be duplicates which is visible from the original solution, i.e. from this line that iterates at least once which implies the key may be overridden:

.forEach(code -> keyMap.put(code, messageKeyRegistry))

So what I suggest is rather group mapping using Collectors.groupingBy resulting in Map<MessageKey, List<MessageKeyRegistry>>:

Map<MessageKey, List<MessageKeyRegistry>> keyMap = EnumSet.allOf(MessageKeyRegistry.class)
        .stream()
        .flatMap(messageKeyRegistry -> messageKeyRegistry.getCodeToMessage().keySet()
            .stream()
            .map(code -> new AbstractMap.SimpleEntry<>(code, messageKeyRegistry)))
        .collect(Collectors.groupingBy(Entry::getKey));

For immutable map, if you mean a map that is read-only use the Collections.unmodifiableMap() wrapper:

Map<MessageKey, List<MessageKeyRegistry>> unmodifiableKeyMap = Collections.unmodifiableMap(keyMap);
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4 Comments

The resulting Map should be immutable, per the OP.
To have an immutable map you should use ...Collectors.toUnmodifiableMap(...)
@HadiJ, that method does not exist in Java 8 (which the OP has tagged).
I have forgotten to it while writing the answer. Wrapping into Collections.unmodifiableMap should help. I have added it.
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I'm not sure the immutability concept of functional programming applying to java streams means immutable returning collections. Let's consider more simple case

final List<String> result = Stream.of("1", "2").collect(Collectors.toList());
result.add("3");
System.out.println(result);
System.out.println(result.getClass());

that outputs

[1, 2, 3]
class java.util.ArrayList
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Not the nicest but something like this should do:

Map<MessageKey, MessageKeyRegistry> keyMap = EnumSet.allOf(MessageKeyRegistry.class)
       .stream()
       .flatMap(messageKeyRegistry -> messageKeyRegistry.getCodeToMessage().keySet().stream().map(code -> new Pair(code, messageKeyRegistry))
       .collect(Collectors.toMap(Pair::getKey, Pair::getValue)));
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2 Comments

Where does the Pair class come from?
javafx package, if you have one. Or you can use Map.Entry otherwise. (or create one yourself... it's just class to hold 2 values)

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