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How does the Java compiler choose the runtime type for a parameterized type with multiple bounds?

I would like to understand better what happens when the Java compiler encounters a call to a method like the one below.
<T extends AutoCloseable & Cloneable>
void printType(T... args) {
System.out.println(args.getClass().getComponentType().getSimpleName());
}
// printType() prints "AutoCloseable"
It is clear to me that there is no type <T extends AutoCloseable & Cloneable> at runtime, so the compiler makes the least wrong thing it can do and creates an array with the type of one of the two bounding interfaces, discarding the other one.
Anyway, if the order of the interfaces is switched, the result is still the same.
<T extends Cloneable & AutoCloseable>
void printType(T... args) {
System.out.println(args.getClass().getComponentType().getSimpleName());
}
// printType() prints "AutoCloseable"
This led me to do some more investigation and see what happens when the interfaces change.
It seems to me that the compiler uses some kind of strict order rule to decide which interface is the most important, and the order the interfaces appear in code plays no role.
<T extends AutoCloseable & Runnable> // "AutoCloseable"
<T extends Runnable & AutoCloseable> // "AutoCloseable"
<T extends AutoCloseable & Serializable> // "Serializable"
<T extends Serializable & AutoCloseable> // "Serializable"
<T extends SafeVarargs & Serializable> // "SafeVarargs"
<T extends Serializable & SafeVarargs> // "SafeVarargs"
<T extends Channel & SafeVarargs> // "Channel"
<T extends SafeVarargs & Channel> // "Channel"
<T extends AutoCloseable & Channel & Cloneable & SafeVarargs> // "Channel"
Question:
How does the Java compiler determine the component type of a varargs array of a parameterized type when there are multiple bounds?
I'm not even sure if the JLS says anything about this, and none of the information I found by googling covers this particular topic.

Typically, when the compiler encounters a call to a parameterised method, it can infers the type (JSL 18.5.2) and can create a correctly typed vararg array in the caller.
The rules are mostly technical ways of saying "find all possible input types and check them" (cases like void, ternary operator, or lambda).
The rest is common sense, such as using the most specific common base class (JSL 4.10.4).
Example:
public class Test {
private static class A implements AutoCloseable, Runnable {
#Override public void close () throws Exception {}
#Override public void run () {} }
private static class B implements AutoCloseable, Runnable {
#Override public void close () throws Exception {}
#Override public void run () {} }
private static class C extends B {}
private static <T extends AutoCloseable & Runnable> void printType( T... args ) {
System.out.println( args.getClass().getComponentType().getSimpleName() );
}
public static void main( String[] args ) {
printType( new A() ); // A[] created here
printType( new B(), new B() ); // B[] created here
printType( new B(), new C() ); // B[] which is the common base class
printType( new A(), new B() ); // AutoCloseable[] - well...
printType(); // AutoCloseable[] - same as above
}
}
JSL 18.2 dictates how to process the constrains for type inference, such as AutoCloseable & Channel is reduced to just Channel.
But the rules do not help answer this question.
Getting AutoCloseable[] from the call may look weird, of course, because we can't do that with Java code.
But in reality the actual type doesn't matter.
At the language level, args is T[], where T is a "virtual type" that is both A and B (JSL 4.9).
The compiler just needs to make sure its usages meet all constrains, and then it knows the logic is sound and there will be no type error (this is how Java generic is designed).
Of course the compiler still need to make a real array, and for the purpose it creates a "generic array".
Thus the warning "unchecked generic array creation" (JLS 15.12.4.2).
In other words, as long as you pass in only AutoCloseable & Runnable, and calls only Object, AutoCloseable, and Runnable methods in printType, the actual array type does not matter.
In fact, printType's bytecodes would be the same, regardless of what kind of array is passed in.
Since printType doesn't care the vararg array type, getComponentType() doesn't and shouldn't matter.
If you want to get the interfaces, try getGenericInterfaces() which returns an array.
Because of type erasure (JSL 4.6), the order of interfaces of T does affect (JSL 13.1) compiled method signature and bytecode. The first interface AutoClosable will be used, e.g. no type check will be done when AutoClosable.close() is called in printType.
But this is unrelated with type interference of method calls of the question, i.e. why AutoClosable[] is created and passed. Many type safeties are checked before erasure, thus the order does not affect type safety. This I think is part of what JSL means by "The order of types... is only significant in that the erasure ... is determined by the first type" (JSL 4.4). It means the order is otherwise insignificant.
Regardless, this erasure rule does cause corner cases such as adding printType(AutoCloseable[]) triggers compile error, when adding printType( Runnable[]) does not. I believe this is an unexpected side effect and is really out of scope.
P.S. Digging too deep may cause insanity, considering that I think I am Ovis aries, view source into assembly, and struggles to answer in English instead of J̶́S͡L̴̀. My sanity score is b҉ȩyon̨d͝ r̨̡͝e̛a̕l̵ numb͟ers͡. T͉͎̫͠u͍r̟̦͝n̪͓͓̭̯̕ ̱̱̞̠̬ͅb̯̠̞̩͎a̘̜̯c̠̮k. ̠̝͕b̭̳͠͡ͅẹ̡̬̦̙f͓͉̼̻o̼͕̱͎̬̟̪r҉͏̛̣̼͙͍͍̠̫͙ȩ̵̮̟̱̫͚ ̢͚̭̹̳̣̩̱͠..t̷҉̛̫͔͉̥͎̬ò̢̱̪͉̲͎͜o̭͈̩̖̭̬.. ̮̘̯̗l̷̞͍͙̻̻͙̯̣͈̳͓͇a̸̢̢̰͓͓̪̳͉̯͉̼͝͝t̛̥̪̣̹̬͔̖͙̬̩̝̰͕̖̮̰̗͓̕͢ę̴̹̯̟͉̲͔͉̳̲̣͝͞.̬͖͖͇͈̤̼͖́͘͢.͏̪̱̝̠̯̬͍̘̣̩͉̯̹̼͟͟͠.̨͠҉̬̘̹ͅ

This is a very interesting question. The relevant part of the specification is §15.12.4.2. Evaluate Arguments:
If the method being invoked is a variable arity method m, it necessarily has n > 0 formal parameters. The final formal parameter of m necessarily has type T[] for some T, and m is necessarily being invoked with k ≥ 0 actual argument expressions.
If m is being invoked with k ≠ n actual argument expressions, or, if m is being invoked with k = n actual argument expressions and the type of the k'th argument expression is not assignment compatible with T[], then the argument list (e1, ..., en-1, en, ..., ek) is evaluated as if it were written as (e1, ..., en-1, new |T[]| { en, ..., ek }), where |T[]| denotes the erasure (§4.6) of T[].
It’s interestingly vague about what “some T” actually is. The simplest and most straight-forward solution would be the declared parameter type of the invoked method; that would be assignment compatible and there is no actual advantage of using a different type. But, as we know, javac doesn’t go that route and uses some sort of common base type of all arguments or picks some of the bounds according to some unknown rule for the array’s element type. Nowadays you might even find some applications in the wild relying on this behavior, assuming to get some information about the actual T at runtime by inspecting the array type.
This leads to some interesting consequences:
static AutoCloseable[] ARR1;
static Serializable[] ARR2;
static <T extends AutoCloseable & Serializable> void method(T... args) {
ARR1 = args;
ARR2 = args;
}
public static void main(String[] args) throws Exception {
method(null, null);
ARR2[0] = "foo";
ARR1[0].close();
}
javac decides to create an array of the actual type Serializable[] here, despite the method’s parameter type is AutoClosable[] after applying type erasure, which is the reason why the assignment of a String is possible at runtime. So it will only fail at the last statement, when attempting to invoke the close() method on it with
Exception in thread "main" java.lang.IncompatibleClassChangeError: Class java.lang.String does not implement the requested interface java.lang.AutoCloseable
It’s blaming the class String here, though we could have put any Serializable object into the array as the actual issue is that a static field of the formal declared type AutoCloseable[] refers to an object of the actual type Serializable[].
Though it is a specific behavior of the HotSpot JVM that we ever got this far, as its verifier does not check assignments when interface types are involved (including arrays of interface types) but defers the check whether the actual class implements the interface to the last possible moment, when trying to actually invoke an interface method on it.
Interestingly, type casts are strict, when they appear in the class file:
static <T extends AutoCloseable & Serializable> void method(T... args) {
AutoCloseable[] a = (AutoCloseable[])args; // actually removed by the compiler
a = (AutoCloseable[])(Object)args; // fails at runtime
}
public static void main(String[] args) throws Exception {
method();
}
While javac’s decision for Serializable[] in the above example seems arbitrary, it should be clear that regardless of which type it chooses, one of the field assignments would only be possible in a JVM with lax type checking. We could also highlight the more fundamental nature of the problem:
// erased to method1(AutoCloseable[])
static <T extends AutoCloseable & Serializable> void method1(T... args) {
method2(args); // valid according to generic types
}
// erased to method2(Serializable[])
static <T extends Serializable & AutoCloseable> void method2(T... args) {
}
public static void main(String[] args) throws Exception {
// whatever array type the compiler picks, it would violate one of the erased types
method1();
}
While this doesn’t actually answer the question what actual rule javac uses (besides that it uses “some T”), it emphasizes the importance of treating arrays created for varargs parameter as intended: a temporary storage (don’t assign to fields) of an arbitrary type you better don’t care about.

Java generics - wildcards extending multiple types [duplicate]

I know there's all sorts of counter-intuitive properties of Java's generic types. Here's one in particular that I don't understand, and which I'm hoping someone can explain to me. When specifying a type parameter for a class or interface, you can bound it so that it must implement multiple interfaces with public class Foo<T extends InterfaceA & InterfaceB>. However, if you're instantiating an actual object, this doesn't work anymore. List<? extends InterfaceA> is fine, but List<? extends InterfaceA & InterfaceB> fails to compile. Consider the following complete snippet:
import java.util.List;
public class Test {
static interface A {
public int getSomething();
}
static interface B {
public int getSomethingElse();
}
static class AandB implements A, B {
public int getSomething() { return 1; }
public int getSomethingElse() { return 2; }
}
// Notice the multiple bounds here. This works.
static class AandBList<T extends A & B> {
List<T> list;
public List<T> getList() { return list; }
}
public static void main(String [] args) {
AandBList<AandB> foo = new AandBList<AandB>(); // This works fine!
foo.getList().add(new AandB());
List<? extends A> bar = new LinkedList<AandB>(); // This is fine too
// This last one fails to compile!
List<? extends A & B> foobar = new LinkedList<AandB>();
}
}
It seems the semantics of bar should be well-defined -- I can't think of any loss of type-safety by allowing an intersection of two types rather than just one. I'm sure there's an explanation though. Does anyone know what it is?

Interestingly, interface java.lang.reflect.WildcardType looks like it supports both upper bounds and lower bounds for a wildcard arg; and each can contain multiple bounds
Type[] getUpperBounds();
Type[] getLowerBounds();
This is way beyond what the language allows. There's a hidden comment in the source code
// one or many? Up to language spec; currently only one, but this API
// allows for generalization.
The author of the interface seems to consider that this is an accidental limitation.
The canned answer to your question is, generics is already too complicated as it is; adding more complexity might prove to be the last straw.
To allow a wildcard to have multiple upper bounds, one has to scan through the spec and make sure the entire system still works.
One trouble I know would be in the type inference. The current inference rules simply can't deal with intersection types. There's no rule to reduce a constraint A&B << C. If we reduced it to
A<<C
or
A<<B
any current inference engine has to go through major overhaul to allow such bifurcation. But the real serious problem is, this allows multiple solutions, but there's no justification to prefer one over another.
However, inference is not essential to type safety; we can simply refuse to infer in this case, and ask programmer to explicitly fill in type arguments. Therefore, difficulty in inference is not a strong argument against intercection types.

From the Java Language Specification:
4.9 Intersection Types
An intersection type takes the form T1 & ... & Tn, n>0, where Ti, 1in, are type expressions. Intersection types arise in the processes of capture conversion (§5.1.10) and type inference (§15.12.2.7). It is not possible to write an intersection type directly as part of a program; no syntax supports this. The values of an intersection type are those objects that are values of all of the types Ti, for 1in.
So why is this not supported? My guess is, what should you do with such a thing? - let's suppose it were possible:
List<? extends A & B> list = ...
Then what should
list.get(0);
return? There's no syntax to capture a return value of A & B. Adding something into such a list would not be possible either, so it's basically useless.

No problem... just declare the type you need in the method signature.
This compiles:
public static <T extends A & B> void main(String[] args) throws Exception
{
AandBList<AandB> foo = new AandBList<AandB>(); // This works fine!
foo.getList().add(new AandB());
List<? extends A> bar = new LinkedList<AandB>(); // This is fine too
List<T> foobar = new LinkedList<T>(); // This compiles!
}

Good question. It took me a while to figure out.
Lets simplify your case: You are trying to do the same as if you declare a class that extends 2 interfaces, and then a variable that has as a type those 2 interfaces, something like this:
class MyClass implements Int1, Int2 { }
Int1 & Int2 variable = new MyClass()
Of course, illegal.
And this is equivalent to what you try to do with generics.
What you are trying to do is:
List<? extends A & B> foobar;
But then, to use foobar, you would need to use a variable of both interfaces this way:
A & B element = foobar.get(0);
Which is not legal in Java. This means, you are declaring the elements of the list as beeing of 2 types simultaneously, and even if our brains can deal with it, Java language cannot.

For what it's worth: if anyone's wondering this because they would truly like to use this in practice, I've worked around it by defining an interface that contains the union of all methods in all the interfaces and class that I'm working with. i.e. I was trying to do the following:
class A {}
interface B {}
List<? extends A & B> list;
which is illegal - so instead I did this:
class A {
<A methods>
}
interface B {
<B methods>
}
interface C {
<A methods>
<B methods>
}
List<C> list;
This still isn't as useful as being able to type something as List<? extends A implements B>, e.g. if someone adds or removes methods to A or B, the typing of the list will not be updated automatically, it requires a manual change to C. But it's worked for my needs.

Java parametrized more then one class. [duplicate]

I know there's all sorts of counter-intuitive properties of Java's generic types. Here's one in particular that I don't understand, and which I'm hoping someone can explain to me. When specifying a type parameter for a class or interface, you can bound it so that it must implement multiple interfaces with public class Foo<T extends InterfaceA & InterfaceB>. However, if you're instantiating an actual object, this doesn't work anymore. List<? extends InterfaceA> is fine, but List<? extends InterfaceA & InterfaceB> fails to compile. Consider the following complete snippet:
import java.util.List;
public class Test {
static interface A {
public int getSomething();
}
static interface B {
public int getSomethingElse();
}
static class AandB implements A, B {
public int getSomething() { return 1; }
public int getSomethingElse() { return 2; }
}
// Notice the multiple bounds here. This works.
static class AandBList<T extends A & B> {
List<T> list;
public List<T> getList() { return list; }
}
public static void main(String [] args) {
AandBList<AandB> foo = new AandBList<AandB>(); // This works fine!
foo.getList().add(new AandB());
List<? extends A> bar = new LinkedList<AandB>(); // This is fine too
// This last one fails to compile!
List<? extends A & B> foobar = new LinkedList<AandB>();
}
}
It seems the semantics of bar should be well-defined -- I can't think of any loss of type-safety by allowing an intersection of two types rather than just one. I'm sure there's an explanation though. Does anyone know what it is?

Interestingly, interface java.lang.reflect.WildcardType looks like it supports both upper bounds and lower bounds for a wildcard arg; and each can contain multiple bounds
Type[] getUpperBounds();
Type[] getLowerBounds();
This is way beyond what the language allows. There's a hidden comment in the source code
// one or many? Up to language spec; currently only one, but this API
// allows for generalization.
The author of the interface seems to consider that this is an accidental limitation.
The canned answer to your question is, generics is already too complicated as it is; adding more complexity might prove to be the last straw.
To allow a wildcard to have multiple upper bounds, one has to scan through the spec and make sure the entire system still works.
One trouble I know would be in the type inference. The current inference rules simply can't deal with intersection types. There's no rule to reduce a constraint A&B << C. If we reduced it to
A<<C
or
A<<B
any current inference engine has to go through major overhaul to allow such bifurcation. But the real serious problem is, this allows multiple solutions, but there's no justification to prefer one over another.
However, inference is not essential to type safety; we can simply refuse to infer in this case, and ask programmer to explicitly fill in type arguments. Therefore, difficulty in inference is not a strong argument against intercection types.

From the Java Language Specification:
4.9 Intersection Types
An intersection type takes the form T1 & ... & Tn, n>0, where Ti, 1in, are type expressions. Intersection types arise in the processes of capture conversion (§5.1.10) and type inference (§15.12.2.7). It is not possible to write an intersection type directly as part of a program; no syntax supports this. The values of an intersection type are those objects that are values of all of the types Ti, for 1in.
So why is this not supported? My guess is, what should you do with such a thing? - let's suppose it were possible:
List<? extends A & B> list = ...
Then what should
list.get(0);
return? There's no syntax to capture a return value of A & B. Adding something into such a list would not be possible either, so it's basically useless.

No problem... just declare the type you need in the method signature.
This compiles:
public static <T extends A & B> void main(String[] args) throws Exception
{
AandBList<AandB> foo = new AandBList<AandB>(); // This works fine!
foo.getList().add(new AandB());
List<? extends A> bar = new LinkedList<AandB>(); // This is fine too
List<T> foobar = new LinkedList<T>(); // This compiles!
}

Good question. It took me a while to figure out.
Lets simplify your case: You are trying to do the same as if you declare a class that extends 2 interfaces, and then a variable that has as a type those 2 interfaces, something like this:
class MyClass implements Int1, Int2 { }
Int1 & Int2 variable = new MyClass()
Of course, illegal.
And this is equivalent to what you try to do with generics.
What you are trying to do is:
List<? extends A & B> foobar;
But then, to use foobar, you would need to use a variable of both interfaces this way:
A & B element = foobar.get(0);
Which is not legal in Java. This means, you are declaring the elements of the list as beeing of 2 types simultaneously, and even if our brains can deal with it, Java language cannot.

For what it's worth: if anyone's wondering this because they would truly like to use this in practice, I've worked around it by defining an interface that contains the union of all methods in all the interfaces and class that I'm working with. i.e. I was trying to do the following:
class A {}
interface B {}
List<? extends A & B> list;
which is illegal - so instead I did this:
class A {
<A methods>
}
interface B {
<B methods>
}
interface C {
<A methods>
<B methods>
}
List<C> list;
This still isn't as useful as being able to type something as List<? extends A implements B>, e.g. if someone adds or removes methods to A or B, the typing of the list will not be updated automatically, it requires a manual change to C. But it's worked for my needs.

Diamond in Generics Java 1.7 - how to write this for Java Compiler in 1.6

how can I write Java 1.7 code for a Java 1.6 compiler, where the diamond can not be used?
Example:
private ReplacableTree<E> convertToIntended(Tree<? extends E> from,ReplacableTree<E> to) {
TreeIterator<? extends E> it = new TreeIterator<>(from.getRoot());
while(it.hasNext()) {
E e = it.next().getElem();
to.add(e);
}
return to;
}
public class TreeIterator<E> implements TreeIter<Node<E>> {
....
}
It is not allowed to write...
TreeIterator<? extends E> it = new TreeIterator<?>(from.getRoot());
TreeIterator<? extends E> it = new TreeIterator<E>(from.getRoot());
TreeIterator<? extends E> it = new TreeIterator<? extends E>(from.getRoot());
Especially the third one is confusing for me. Why doesn't it work? I just want to read Elements from a Tree (which could be a subtype tree), and when puch each of it in a new Tree with Elements of type E.

Wildcard types are not permitted as type arguments in class instance creation expressions:
It is a compile-time error if any of the type arguments used in a class instance creation expression are wildcard type arguments (§4.5.1).
so the first and third variants are not valid.
Variant 2 is invalid because the TreeIterator<E> constructor wants a Node<E>, but you give it a Node<? extends E>.
As for the solution, Java 5 and 6 did not have type inference for constructors, but do have type inference for methods, and in particular capture conversion. The following ought to compile:
TreeIterator<? extends E> it = makeIterator(from.getRoot());
where
private <E> TreeIterator<E> makeIterator(Node<E> node) {
return new TreeIterator<E>(node);
}
Edit: You asked in the comment:
The contstructor parameter type for TreeIterator is Node<E>. The constructor parameter of Node<E> therefore is E. When writing variant two, eclipse says the following: The constructor TreeIterator<E>(Node<capture#2-of ? extends E> ) is undefined What does that mean?
Being a wildcard type, the type Node<? extends E> represents a family of types. Node<capture#2-of ? extends E> refers to a specific type in that family. That distinction is irrelevant in this case. What matters is that Node<? extends E> is not a subtype of Node<E>, and hence you can't pass an instance of Node<? extends E> to a constructor expecting a Node<E>.

In short you don't write Java 7 code for a Java 6 compiler - you have to use the old, duplicative non-diamond syntax. And no, you can't specify a target of 1.6 with source 1.7, it won't work!

meriton already explained it well. I just want to suggest that you could as well do it without the wildcard declaration:
TreeIterator<E> it = new TreeIterator<E>(from.getRoot());

Usually, <> means to just use the same type parameter as in the declaration to the left. But in this case, that declaration is a wildcard.
It doesn't make sense to make a constructor with a wildcard type parameter, new TreeIterator<? extends E>(...) because, usually, if you don't care what parameter to use, you should just pick any type that satisfies that bound; which could be E, or any subtype thereof.
However, in this case, that doesn't work because the constructor of TreeIterator<E> takes an object with the type parameter <E>. You didn't show the source code of TreeIterator, so I can't see what it does, but chances are that its bound is too strict. It could probably be refactored to make the type parameter <? extends E>.
But there are some cases where that is not possible. In such a case, you can still eliminate the need for the type parameter E through a "capture helper" (what meriton suggests above) to turn something which takes parameter E into something that takes a wildcard ? extends E.

I know this is an old question, but in case someone stumbles on this, I would have thought the most obvious way of writing it would have been:
private <U extends E> ReplaceableTree<E> convertToIntended(Tree<U> from, ReplaceableTree<E> to)
{
TreeIterator<U> it = new TreeIterator<U>(from.getRoot());
while(it.hasNext())
{
E e = it.next().getElem();
to.add(e);
}
return to;
}
I don't think such a change would break existing code as the type constraints are the same from the existing signature to this one.

Why can't you have multiple interfaces in a bounded wildcard generic?

I know there's all sorts of counter-intuitive properties of Java's generic types. Here's one in particular that I don't understand, and which I'm hoping someone can explain to me. When specifying a type parameter for a class or interface, you can bound it so that it must implement multiple interfaces with public class Foo<T extends InterfaceA & InterfaceB>. However, if you're instantiating an actual object, this doesn't work anymore. List<? extends InterfaceA> is fine, but List<? extends InterfaceA & InterfaceB> fails to compile. Consider the following complete snippet:
import java.util.List;
public class Test {
static interface A {
public int getSomething();
}
static interface B {
public int getSomethingElse();
}
static class AandB implements A, B {
public int getSomething() { return 1; }
public int getSomethingElse() { return 2; }
}
// Notice the multiple bounds here. This works.
static class AandBList<T extends A & B> {
List<T> list;
public List<T> getList() { return list; }
}
public static void main(String [] args) {
AandBList<AandB> foo = new AandBList<AandB>(); // This works fine!
foo.getList().add(new AandB());
List<? extends A> bar = new LinkedList<AandB>(); // This is fine too
// This last one fails to compile!
List<? extends A & B> foobar = new LinkedList<AandB>();
}
}
It seems the semantics of bar should be well-defined -- I can't think of any loss of type-safety by allowing an intersection of two types rather than just one. I'm sure there's an explanation though. Does anyone know what it is?

Interestingly, interface java.lang.reflect.WildcardType looks like it supports both upper bounds and lower bounds for a wildcard arg; and each can contain multiple bounds
Type[] getUpperBounds();
Type[] getLowerBounds();
This is way beyond what the language allows. There's a hidden comment in the source code
// one or many? Up to language spec; currently only one, but this API
// allows for generalization.
The author of the interface seems to consider that this is an accidental limitation.
The canned answer to your question is, generics is already too complicated as it is; adding more complexity might prove to be the last straw.
To allow a wildcard to have multiple upper bounds, one has to scan through the spec and make sure the entire system still works.
One trouble I know would be in the type inference. The current inference rules simply can't deal with intersection types. There's no rule to reduce a constraint A&B << C. If we reduced it to
A<<C
or
A<<B
any current inference engine has to go through major overhaul to allow such bifurcation. But the real serious problem is, this allows multiple solutions, but there's no justification to prefer one over another.
However, inference is not essential to type safety; we can simply refuse to infer in this case, and ask programmer to explicitly fill in type arguments. Therefore, difficulty in inference is not a strong argument against intercection types.

From the Java Language Specification:
4.9 Intersection Types
An intersection type takes the form T1 & ... & Tn, n>0, where Ti, 1in, are type expressions. Intersection types arise in the processes of capture conversion (§5.1.10) and type inference (§15.12.2.7). It is not possible to write an intersection type directly as part of a program; no syntax supports this. The values of an intersection type are those objects that are values of all of the types Ti, for 1in.
So why is this not supported? My guess is, what should you do with such a thing? - let's suppose it were possible:
List<? extends A & B> list = ...
Then what should
list.get(0);
return? There's no syntax to capture a return value of A & B. Adding something into such a list would not be possible either, so it's basically useless.

No problem... just declare the type you need in the method signature.
This compiles:
public static <T extends A & B> void main(String[] args) throws Exception
{
AandBList<AandB> foo = new AandBList<AandB>(); // This works fine!
foo.getList().add(new AandB());
List<? extends A> bar = new LinkedList<AandB>(); // This is fine too
List<T> foobar = new LinkedList<T>(); // This compiles!
}

Good question. It took me a while to figure out.
Lets simplify your case: You are trying to do the same as if you declare a class that extends 2 interfaces, and then a variable that has as a type those 2 interfaces, something like this:
class MyClass implements Int1, Int2 { }
Int1 & Int2 variable = new MyClass()
Of course, illegal.
And this is equivalent to what you try to do with generics.
What you are trying to do is:
List<? extends A & B> foobar;
But then, to use foobar, you would need to use a variable of both interfaces this way:
A & B element = foobar.get(0);
Which is not legal in Java. This means, you are declaring the elements of the list as beeing of 2 types simultaneously, and even if our brains can deal with it, Java language cannot.

For what it's worth: if anyone's wondering this because they would truly like to use this in practice, I've worked around it by defining an interface that contains the union of all methods in all the interfaces and class that I'm working with. i.e. I was trying to do the following:
class A {}
interface B {}
List<? extends A & B> list;
which is illegal - so instead I did this:
class A {
<A methods>
}
interface B {
<B methods>
}
interface C {
<A methods>
<B methods>
}
List<C> list;
This still isn't as useful as being able to type something as List<? extends A implements B>, e.g. if someone adds or removes methods to A or B, the typing of the list will not be updated automatically, it requires a manual change to C. But it's worked for my needs.