We're just wrapping up the implementation of Ceylon's enumerated types (algebraic types, if you prefer that term). This feature is turning out significantly nicer than what was originally defined in the spec.
Unions and switch
The simplest kind of enumerated type is simply a union of classes. For example, the type Integer|Float is a very simple enumerated type. Since Float and Integer are distinct classes and neither inherits from the other, the compiler reasons that the types are disjoint, meaning they have no common instances. Therefore, it lets us write the following:
void add(Integer|Float x) {
switch (x)
case (is Integer x) { integers.add(x); }
case (is Float x) { floats.add(x); }
}
Notice that inside the block that follows a case, the type of the switched variable is narrowed to the case type. We know that x is an Integer in the first case, and a Float in the second, so the compiler lets us make use of this additional information.
There are two basic things to appreciate about Ceylon's switch construct:
- The
cases must be disjoint. The behavior never depends upon the order in which the cases appear.
- If the
cases don't completely cover all possible values of the switch expression type, we need to include an else clause.
A switch statement which covers all possible values of the switched type is said to have an exhaustive list of cases.
The purpose of the required else clause is to syntactically distinguish switch statements which don't need to be fixed and recompiled when a new type is added.
void add(Integer|Float|String x) {
switch (x)
case (is Integer x) { integers.add(x); }
case (is Float x) { floats.add(x); }
else {}
}
Enumerated classes
We can extend this idea for an interface or abstract class. Ceylon lets us declare that an abstract class or interface is equivalent to a union using the of clause. For example, the root class of the type hierarchy, Void, is declared like this:
shared abstract class Void() of Nothing | Object {}
This says that the class Void has exactly the same instances as the union type Nothing|Object. So when we encounter a value of type Void, we know it must either be an instance of Object, or an instance of Nothing. We can write:
void printVoid(Void v) {
switch (v)
case (is Nothing) { print("nothing"); }
case (is Object) { print(v); }
}
The compiler won't let us define a third subclass of Void that doesn't extend either Nothing or Object. We call the types mentioned in the of clause the cases of an enumerated type.
Since an object declaration is also a type, we can even include the names of objects in the of clause. For example, this is how the language module expressed the fact that the only instance of Nothing is the null value:
shared abstract class Nothing() of null {}
shared object null extends Nothing() {}
More commonly, there will be multiple values, allowing us to recapture the semantics of a Java enum, without introducing a special language construct. For example, Ceylon's Boolean is an enumerated type with two values:
shared abstract class Boolean() of true | false {}
shared object true extends Boolean() {
shared actual String string { return "true"; }
}
shared object false extends Boolean() {
shared actual String string { return "false"; }
}
Therefore, we can switch on a Boolean value:
void printBoolean(Boolean bool) {
switch (bool)
case (true) { print("true"); }
case (false) { print("false"); }
}
Note that the classic example of a Java enumerated type works out significantly more verbose in Ceylon:
shared abstract class Suit()
of hearts|diamonds|clubs|spades {}
shared object hearts extends Suit() {}
shared object diamonds extends Suit() {}
shared object clubs extends Suit() {}
shared object spades extends Suit() {}
I think that's OK. The tradeoff is that enumerated types in Ceylon are simply much more powerful and flexible.
Enumerated interfaces
Enumerated interfaces are a little more special. Consider:
interface Association
of OneToOne | OneToMany | ManyToOne | ManyToMany { ... }
Interfaces support multiple inheritance, so in general interface types are not disjoint. But when an interface declares an enumerated list of subinterfaces, the compiler enforces that those interfaces be disjoint by rejecting any attempt to define a concrete class or object that is a subtype of more than one of them. This code is not well-typed:
class BrokenAssociation()
satisfies OneToOne & OneToMany {} //error: inherits multiple cases of enumerated type
Note that the types named in the of clause don't need to be direct subtypes of the enumerated type. The following code is well-typed:
interface ToOne satisfies Association {}
class OneToOne() satisfies ToOne {}
class ManyToOne() satisfies ToOne {}
The interface ToOne is a subtype of the enumerated type Association but not of any of its cases. That's acceptable, because ToOne is an abstract type that can't be directly instantiated. Of course, every one of its concrete subtypes must be a subtype of one of the enumerated cases of Association.
Handling multiple cases at once
A case of a switch can handle more than one type at once. For example, ToOne could be a case type:
Association assoc = ... ;
switch (assoc)
case (is ToOne) { ... }
else { ... }
Even better, we can form unions of the cases of the enumerated type:
Association assoc = ... ;
switch (assoc)
case (is OneToOne|ManyToOne) { ... }
case (is OneToMany|ManyToMany) { ... }
For object cases, the syntax is slightly different:
switch (suit)
case (hearts, diamonds) { return red; }
case (clubs, spades) { return black; }
How the compiler reasons about enumerated types
Now let's see something cool. When the compiler encounters an enumerated type in a switch statement, it:
- first expands the type to a union of its cases. And it does this recursively to any of the cases which are themselves enumerated types, and then
- to determine if a
switch statement has an exhaustive list of case branches, takes the union of each of the case types and determines if it is a supertype of the union type produced in step 1.
Thus, the following switch is well-typed:
void printVoid(Void v) {
switch (v)
case (nothing) { print("nothing"); }
case (is Object) { print(v); }
}
Likewise, remembering that Boolean? means Nothing|Boolean, we can write:
void printBoolean(Boolean? bool) {
switch (bool)
case (true) { print("true"); }
case (false) { print("false"); }
case (nothing) { print("nothing"); }
}
But what about disjointness of the case types? How is that determined? Well, first of all, the compiler is able to reason that the intersection of two types is empty if they inherit from different cases of an enumerated type. So whenever the compiler encounters a type like Nothing&Boolean, it automatically simplifies the type to Bottom.
So the disjointness restriction on cases of a switch boils down to checking that for each pair of cases, the intersection of the case types is Bottom.
