generics - Java API class having a method with a meaningless type wildcard -


in java 8, class java.util.optional (javadoc) class offers functionality of "maybe" monad or option type.

more directly:

public final class java.util.optional<t> extends object

a container object may or may not contain non-null value. if value present, ispresent() return true , get() return value.

one of methods is:

<u> optional<u> map(function<? super t,? extends u> mapper)

if value present, apply provided mapping function it, , if result non-null, return optional describing result.

the question why map() use type wildcard on u type.

my understanding of type wildcarding that:

enter image description here

? super t designates type set of classes given subclassing path object t (both object , t included) union set of interfaces found on subinterfacing path pulled in "as sidedish" via implements.

and ? extends u designates type set of classes extend u (uincluded).

so 1 write:

<u> optional<u> map(function<? super t,u> mapper)

without loss of information.

or not?

not quite.

a function<? super t, u> not same function<? super t, ? extends u>.

for example, still optional<charsequence>, if passed function<object, string> method. if method defined <u> optional<u> map(function<? super t, u> mapper), not possible.

that because generics invariant: <t> not same <? extends t>. that's design decision implemented in java language.

let's see how jon skeet explains happen if generics not invariant:

class animal { }  class dog extends animal { }  class cat extends animal { } 
public void ouch() {     list<dog> dogs = arrays.aslist(new dog(), new dog());     list<animal> animals;     // legal, right? because list of dogs list of animals.     list<animal> animals = dogs;     // legal, right? because cat added     // list of animals, because cat animal.     animals.add(new cat());     // unfortunately, have confused cat. }

although i'm not entirely sure mean in comments, i'll try elaborate.

if have full control on function provide, doesn't matter whether method's signature function<? super t, u> or function<? super t, ? extends u>, you'll adapt function accordingly. authors of method wanted method as flexible possible, allowing 1 provide function second parameter subclass of u, instead of u itself. widen lower bound u subtype of u.

so function should read <u> optional<? the-most-general-but-fixed-supertype-of u> map(function<? super t, u> mapper) expressing way awkward.

i indeed awkward. in addition, there difference between proposed notation , actual method signature of map(), involve implications of lower bounds , upper bounds.

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