How to properly use the combination of .map and .orElseGet within stream API?

Question

I have the following code:

private List<User> findUsers(...) {
...

    return usersData.stream() //userData is another list of objects
        .findFirst()
        .map(this::getCode)
        .map(code-> {
            if (...) {
                Optional<AnotherObject> anotherObject = getAnotherObject();
                return anotherObject.map(userService::getUsersFromAnotherObject) // getUsersFromAnotherObject returns List<User> but the whole line returns Optional of List<User>
            } else {
                ...
                return null;
            }
        }).orElseGet(() -> findXYZ(...));
}

which does not compile and says:
"Bad return type in lambda expression: List<User> cannot be converted to Optional<List<User>>"
even though findXYZ and all other if/else statements return in fact type List.

Could anybody explain to me what is wrong with the code?

EDIT: Sorry, I noticed that one of if statements is actually returning the Optional of List

If anybody is interested, I solved it simply editing the first "if" to:

return userService.getUsersFromAnotherObject(anotherObject.orElse(null));

what does those `...` implement and `return` and what is the return type of `getCode` and `findXYZ`? — Naman, Jul 16 '20 at 18:15
he mentioned that the return type of those 'return' and findXyz is List — lorraine batol, Jul 16 '20 at 18:22
I’d find it easier to help if you gave us [a reproducible example](https://stackoverflow.com/help/minimal-reproducible-example), that is, one that gives that exact error message. — Ole V.V., Jul 16 '20 at 18:40

lorraine batol · Accepted Answer · 2020-07-17T08:41:03.227

1

(edited)

In your case, the return of the two should match because that's the use of the orElseGet() - to give an alternate value of the same type.

.map(code -> return ...)
.orElseGet(() -> return ...)

There are two options in your case:

since your map() returns Optional<List<User>>, you can update the findXyz() to return the same

update code of map() to something like below (return List<User> without Optional wrap, then you can keep your findXyz() in it's original).

     usersData.stream().findFirst()
     .map(this::getCode)
     .map(code-> {
         if (...) {
             Optional<AnotherObject> anotherObject = getAnotherObject();
             Optional<List<User>> optUserList = anotherObject.map(userService::getUsersFromAnotherObject)
             return optUserList.isPresent() ? optUserList.get() : null;
         } else {
                 ...
             return null;
         }
     }).orElseGet(() -> findXYZ(...));`

edited Jul 17 '20 at 08:41

answered Jul 16 '20 at 17:56

lorraine batol

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Unfortunately I cannot use second "findFirst()". There is no such method to choose after the brackets. – Peters_ Jul 17 '20 at 07:10
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Oops sorry, i missed that you already have a findFirst before map() – lorraine batol Jul 17 '20 at 07:24
1

the output of findXyz() will need to be Optional> too (sorry cannot edit my answer,on mobile phone only) – lorraine batol Jul 17 '20 at 07:45

score 0 · Answer 2 · edited Jul 16 '20 at 18:37

0

Just add a .get() after orElseGet()

private List<User> findUsers(...)
{
    ...
    return usersData.stream() //userData is another list of objects 
           .findFirst()
           .map(this::getCode)
           .map(code-> {
               if (...) {
                   return ...
               }
               else {
                   return ...;
               }
           })
           .orElseGet(() -> findXYZ(...))
           .get(); 
}

Because .orelse get return object of Optional class

edited Jul 16 '20 at 18:37

Ole V.V.

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answered Jul 16 '20 at 17:59

Dhruv Tailor

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is it possible to return the element itself(code) after stream from orElseGet()? – Dev Aug 19 '21 at 15:09

How to properly use the combination of .map and .orElseGet within stream API?

2 Answers2