Why Bar.go
is OK
with argument f2
but not with argument f1
?
public class HelloWorld {
public static void main(String[] args) {
Foo<Foo<?>> f1 = new Foo<Foo<?>>();
Foo<Foo<String>> f2 = new Foo<Foo<String>>();
Bar.go(f1); // not OK
Bar.go(f2); // OK
}
public static void p(Object o) {
System.out.println(o);
}
}
class Foo<E> {
}
class Bar {
public static <T> void go(Foo<Foo<T>> f) {
}
}
Shouldn't the compiler automatically infer type T
as capture of ?
in both cases?
Foo<Foo<?>> f1 = new Foo<Foo<?>>();
This implies that the type is unknown and objects of any type can be added to Foo<Foo<?>>
that are heterogeneous and compiler cannot guarantee that all object in Foo<Foo<?>>
are of same type. Hence it cannot be passed to Bar.go
that takes a bounded type as parameter.
You can instead declare that as Foo<Foo<Object>> f1 = new Foo<Foo<Object>>();
to pass it to Bar.go
where you explicitly mention everything is of type Object
.
Great question!
(In the following comments, wrt a class generic in E
like Foo< E >
, define "covariant method" as a method that returns an E
without having any parameters using E
, and a "contravariant method" as the opposite: one which takes a formal parameter of type E
but doesn't return a type involving E
. [The real definition of these terms is more complicated, but never mind that for now.])
It seems that the compiler is trying to bind T
to Object
in the case of f1
, because if you do
class Bar0 {
public static < T > void go( Foo< Foo< ? extends T > > f ) {
// can pass a Foo< T > to a contravariant method of f;
// can use any result r of any covariant method of f,
// but can't pass T to any contravariant method of r
}
}
then the go(f1)
works, but now go(f2)
doesn't, because even though Foo< String > <: Foo< ? extends String >
Foo< String > <: Foo< ? extends String >
, that does not imply that Foo< Foo< String > > <: Foo< Foo< ? extends String > >
Foo< Foo< String > > <: Foo< Foo< ? extends String > >
.
Here are a few modifications that compile for both f1
and f2
:
class Bar1 {
public static < T > void go( Foo< ? super Foo< T > > f ) {
// can't properly type the results of any covariant method of f,
// but we can pass a Foo< T > to any contravariant method of f
}
}
class Bar2 {
public static < T > void go( Foo< ? extends Foo< ? extends T > > f ) {
// can't pass a Foo< T > to a contravariant method of f;
// can use result r of any covariant method of f;
// can't pass a T to a contravariant method of r;
// can use result of covariant method of r
}
}
A Nice Read What do multi-level wildcards mean?
Example:
Collection< Pair<String,Long> > c1 = new ArrayList<Pair<String,Long>>();
Collection< Pair<String,Long> > c2 = c1; // fine
Collection< Pair<String,?> > c3 = c1; // error
Collection< ? extends Pair<String,?> > c4 = c1; // fine
Of course, we can assign a Collection<Pair<String,Long>>
to a Collection<Pair<String,Long>>
. There is nothing surprising here.
But we can not assign a Collection<Pair<String,Long>>
to a Collection<Pair<String,?>>
. The parameterized type Collection<Pair<String,Long>>
is a homogenous collection of pairs of a String and a Long ; the parameterized type Collection<Pair<String,?>>
is a heterogenous collection of pairs of a String and something of unknown type. The heterogenous Collection<Pair<String,?>>
could for instance contain a Pair<String,Date>
and that clearly does not belong into a Collection<Pair<String,Long>>
. For this reason the assignment is not permitted.
I will prove that if the compiler allows Bar.go(f1);
, the type system(safety) would be broken:
Java grammar allows you to use T
as a type to declare variables in go()
. Something like: T t = <something>
.
Now, let's use ArrayList
instead of Foo
,
Then we have:
class HW {
public static void main(String[] args) {
ArrayList<ArrayList<?>> f1 = new ArrayList<ArrayList<?>>();
go(f1); // not OK
}
public static <T> void go(ArrayList<ArrayList<T>> f) {
}
}
ArrayList<?>
is supertype of ArrayList<String>
, it is also supertype of ArrayList<Integer>
, meaning that you can do the following in main
:
ArrayList<?> s = new ArrayList<String>();
f1.add(s);
ArrayList<?> i = new ArrayList<Integer>();
f1.add(i);
Now, let's assume that the compiler allows you to call go()
with f1
as argument. The option to infer T
are:
T = Object
, but ArrayList<ArrayList<Object>>
is not ArrayList<ArrayList<?>>
because ArrayList<Object>
is not the same type as ArrayList<?>
So that is not allowed option.
T = ?
, then we would be able to do:
public static <T> void go(ArrayList<ArrayList<T>> f) { ArrayList<T> alt1 = f.get(0); // ArrayList<String> T str = alt1.get(0); ArrayList<T> alt2 = f.get(1); // ArrayList<Integer> alt2.add(str); // We have added String to List<Integer> // ... type system broken }
go()
to work in both cases you have to do:
public static void go(ArrayList<? extends ArrayList<?>> f) {
}
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