Expressions - The Java Language Specification - page 472

This follows from the covariant subtype relation among array types. The con-
straint A << F in this case means that A << U [] . A is therefore necessarily an
array type V [] , or a type variable whose upper bound is an array type V [] - oth-
erwise the relation A << U [] could never hold true. It follows that V [] << U [] .
Since array subtyping is covariant, it must be the case that V << U .
• If F has the form G < ..., Y k-1 , U , Y k+1 , ... > , where U is a type expression that in-
volves T j , then if A has a supertype of the form G < ..., X k-1 , V , X k+1 , ... > where V is a
type expression, this algorithm is applied recursively to the constraint V = U .
For simplicity, assume that G takes a single type argument. If the method in-
vocation being examined is to be applicable, it must be the case that A is a
subtype of some invocation of G . Otherwise, A << F would never be true.
In other words, A << F , where F = G < U > , implies that A << G < V > for some V .
Now, since U is a type expression (and therefore, U is not a wildcard type ar-
gument), it must be the case that U = V , by the non-variance of ordinary para-
meterized type invocations.
The formulation above merely generalizes this reasoning to generics with an
arbitrary number of type arguments.
• If F has the form G < ..., Y k-1 , ? extends U , Y k+1 , ... > , where U involves T j , then if A
has a supertype that is one of:
♦ G < ..., X k-1 , V , X k+1 , ... > , where V is a type expression. Then this algorithm is
applied recursively to the constraint V << U .
Again, let's keep things as simple as possible, and consider only the case
where G has a single type argument.
A << F in this case means A << G <? extends U > . As above, it must be the case
that A is a subtype of some invocation of G . However, A may now be a sub-
type of either G < V > , or G <? extends V > , or G <? super V > . We examine these cases
in turn. The first variation is described (generalized to multiple arguments) by
the sub-bullet directly above. We therefore have A = G < V > << G <? extends U > .
The rules of subtyping for wildcards imply that V << U .
♦ G < ..., X k-1 , ? extends V , X k+1 , ... > . Then this algorithm is applied recursively to
the constraint V << U .
Extending the analysis above, we have A = G <? extends V > << G <? extends U > .
The rules of subtyping for wildcards again imply that V << U .
♦ Otherwise, no constraint is implied on T j .