How do I use Java-8's flatMap operation for certain cases?

Question

I am using Java-8 's flatMap operation to make a part of my code smaller.

I know that if my code is as follows:

if (giftOptional.isPresent()) {
    Gift gift = giftOptional.get();
    Optional<Boolean> giftIsWrapped = gift.isWrapped();
    return giftIsWrapped.isPresent() && giftIsWrapped.get();
}
return false;

then it can be changed to:

return giftOptional.flatMap(Gift::isWrapped)
                   .orElse(Boolean.FALSE);

but what if the isWrapped method does not return an Optional? That is, if the code is like this:

if (giftOptional.isPresent()) {
    Gift gift = giftOptional.get();
    boolean giftIsWrapped = gift.isWrapped();
    return giftIsWrapped;
}
return false;

then can the same code apply? That is:

return giftOptional.flatMap(Gift::isWrapped)
                   .orElse(Boolean.FALSE);

Also, what if I need to add some argument to the function? For example, the code is:

if (giftOptional.isPresent()) {
    String gift = giftOptional.get();
    boolean giftIsWrapped = gift.equals("some string");
    return giftIsWrapped;
}
return false;

what should be the corresponding code using flatMap in that case?

Misha · Accepted Answer · 2018-01-29 09:47:14Z

3

You needed to use flatMap to unwrap or "flatten" the nested Optional. If isWrapped returns a boolean, simply use map instead of flatMap:

giftOptional.map(Gift::isWrapped).orElse(false);

You can always use a lambda expression inside map to express the mapping logic. In your case, you can use .map(gift -> gift.equals("some string")). That happens to be the same as "some string"::equals, but you cannot always reduce a lambda into a method reference.

The code above will work with any return type. However, since in your case you are dealing with a boolean, you can use filter instead of map:

giftOptional.filter(Gift::isWrapped).isPresent();

edited Jan 29, 2018 at 9:47

answered Jan 29, 2018 at 4:44

Misha

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1 Comment

Sanjiban Bairagya Over a year ago

Ah, got it, thanks. Can you please also tell me what do I do when I need to add an argument to the function? You can check the last part of my question if you want to look at an example code for reference.

Bohemian · Accepted Answer · 2018-01-29 05:47:04Z

2

You want filter():

return giftOptional.filter(Gift::isWrapped).orElse(Boolean.FALSE);

filter() takes a Predicate, which can be a method reference, and works as you would expect and as you want in your case.

You can still use filter() with a method reference for comparison with a String:

return giftOptional.filter("some string"::equals).orElse(Boolean.FALSE);

edited Jan 29, 2018 at 5:47

answered Jan 29, 2018 at 4:50

Bohemian♦

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7 Comments

Sanjiban Bairagya Over a year ago

Got it, thanks. Would you also happen to know what should be the code if the function required an argument to be passed to it, as mentioned in the final part of my question?

Bohemian Over a year ago

@SanjibanBairagya see edited answer for alternate code

Sanjiban Bairagya Over a year ago

I'm afraid the solutions mentioned here are not compiling. I'm getting the following errors respectively: error: incompatible types: Boolean cannot be converted to Gift, and, error: incompatible types: Boolean cannot be converted to String

Bohemian Over a year ago

@SanjibanBairagya what code exactly is giving the error(s)?

Bohemian Over a year ago

@SanjibanBairagya because Predicate<T> is a Function<T, Boolean>, in this case map() and filter() are effectively the same thing. You should use whatever makes the code clearer,

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