HenryNguyen :
I'm using Java 1.8.0_151 and there is some code that doesn't compile and I don't understand:
Optional optional = Optional.of("dummy");
Optional<Boolean> result1 = optional.map(obj -> true); // works fine
boolean result2 = result1.orElse(false); // works fine
boolean result3 = optional.map(obj -> true).orElse(false); // compilation error: Incompatible types: required boolean, found object
Object result4 = optional.map(obj -> true).orElse(false); // works fine
why it works fine on result1 but gives compilation error on result3?
Additional info:
- In the first line, when I change
OptionaltoOptional<String>, result3 is also able to compile - When I break
result3into 2 lines: likeresult1andresult2,result3is able to compile
Eugene :
Once you lose the type safety - it is lost for chained calls as-well. That is Optional<Object> != Optional. So when you do
Optional optional = Optional.of("dummy");
optional.map()
The map can only accept a raw Function and nothing else, which will return an Object obviously.
The correct way is to add type information :
Optional<String> optional = Optional.of("dummy");
Or you can cast unsafely:
boolean result3 = (boolean) optional.map(obj -> true).orElse(false)
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