Generic parameters constrained by polymorphic types allow unsafe operations but are not widened on return positions

[malibuzios]

TypeScript Version:

1.9.0-dev.20160405

Code

class Animal {
    name: string;
}

class Dog extends Animal {
    bark() {};
}

class Cat extends Animal {
    meow() {};
}

class Box<V> {
    item: V;
}

function f<T extends Box<Animal>>(box: T) {
    box.item = new Cat(); // <-- This is perfectly valid and allowed by the constraint
    return box; // <-- The returned type is T
}

let result = f(new Box<Dog>()); // Type of 'result' is inferred as Box<Dog>
result.item.bark(); // <-- No compilation error. Run-time error..

Expected behavior:
Type of result should be inferred as Box<Animal>.
result.item.bark(); should give a compilation error.

Actual behavior:
Type of result is inferred as Box<Dog>.
result.item.bark(); does not error at compile-time but does error at run-time.

Comments:
Type of box should be widened from T to Box<Animal> within the body of the function, as there is no (current) way to ensure the members of box are not reassigned with types incompatible to the ones specified in T. The return type should therefore be Box<Animal> and not T.

[Edits: rewritten main example, heavily shortened and removed unneeded sections]

[malibuzios]

Another way to state this is that generic constraints, as implemented today, ensure type-safety [for polymorphic types] in the 'read' direction, but not in the 'write' direction. Implicitly making the affected type immutable is a step forward, although I do understand it may be difficult to recursively apply this to all its sub properties etc.. [Edit: thinking about this more I'm not sure it would even be possible in some cases..]

Alternatively, if the type would be widened to the constraint in return positions, it would be similar to the effect of having the parameter type simply set to the constraint:

function f(box: Box<Animal>) {
    box.item = new Cat(); // <-- This is perfectly valid and allowed by the type
    return box; // <-- The returned type is Box<Animal>
}
let result = f(new Box<Dog>()); // Type of 'result' is Box<Animal>

So the benefit of having it in a constraint may be quite limited..

[malibuzios]

In order to "convert" this type to a "safe" one for this context:

interface Example<T> {
    a: T;
    possiblyMutatingMethod(arg: T);
}

one would need to completely remove possiblyMutatingMethod and in general any method that contains the T type in any of its parameters (return types should be fine, I believe, but I'm not 100% sure). However since a may represent a setter and may mutate the object, it needs to be removed as well.. Leaving an empty interface that is quite useless. There is no way to just "mark" T as "immutable"..

---

Essentially what happens here is that it takes the the type Box<Dog> and substitutes it with the "more capable" type Box<Animal> and then substitutes it back to the "less capable" type Box<Dog> on return positions.

In C# this is not permitted. The answer by Eric Lippert cites the Liskov Substitution Principle:

The type system does not permit that conversion because doing so would violate the rule that the capabilities of the source type must not be less than the capabilities of the target type. (This is a form of the famous "Liskov substitution principle".)

[Arnavion]

The keywords you're looking for are covariance / contravariance, i.e. #1394

[malibuzios]

@Arnavion

I don't think this is strictly related to covariance and contravariance. After further research I started to see it is particular to the behavior of the any type. It has to do with the fact that any can be assigned to anything (including number), this causes a generic type that has any as it its type argument to allow unsafe operations:

class Box<T> {
    item: T;
}

let boxContainingNumber = new Box<number>();
let boxContainingAnything = new Box<any>();

boxContainingNumber.item = 42;
boxContainingAnything = boxContainingNumber; 
boxContainingAnything.item = new Date();
boxContainingNumber = boxContainingAnything; // No error :`(

But this is not really the issue here (it is a different issue by itself). The issue is that the semantics of the constraint (e.g T extends Cage<Animal>) should provide the 'guarantee' that type is at least what is specified. However, Example<number> can only hold a number, it cannot hold 'anything' (just like Cage<Tiger> can only hold a tiger, not any animal).

Despite this fact it is still allowed to use it in this position. The fact that no one for more than 4 years has noticed that means that very few people actually understand it and I believe it should be reconsidered.

[Arnavion]

It is absolutely related. To be specific, TS treats types as covariant. Type T is assignable to type U if all members of T are assignable to U, which is the definition of covariance.

It also is not specific to any, as you can see from that issue thread. Eg arrays:

class Animal { foo: string; }
class Cat extends Animal { meow() { } }
class Dog extends Animal { bark() { } }

var cats: Cat[] = [new Cat()];
var animals: Animal[] = cats;
animals[0] = new Dog();
cats[0].meow(); // Blows up at runtime because cats[0] is a Dog

or union types:

var x: { y: number } = { y: 5 };
function f(a: { y: number | string }) {
    a.y = "foo";
}
f(x); // x.y is now "foo"

[T18970237136]

Hi @malibuzios,
if I understand you correctly then you are describing two things.

The first one is that any is assignable to everything, and everything is assignable to any (AFAIK):

let f: any;
let c: number;
f = c = f; // works

This is by design. See the Any type in the Handbook/Spec.

The second is that generic types in Typescript appear to be covariant, just as @Arnavion described.
You are right that such behavior is not allowed in C# by default, but you can explicitely mark type parameters in generic interfaces as covariant or contravariant, if the type parameter is used in methods/properties and follows some rules (e.g. a contravariant type doesn't work as a return type, and a covariant type doesn't work as function argument type), see Variance in Generic Interfaces.

However, in Typescript, Function argument types are considered bivariant as described in the thread that @Arnavion linked, whereas in C# they are considered contravariant (so it makes sense to treat generic type parameters as covariant in TypeScript).

[malibuzios]

@Arnavion

I understand what you are describing, but the particular aspect I'm concentrating on has one 'subtle' difference from the examples you gave:

The programmer is given the 'illusion' that a type is preserved and type-safety is guaranteed when passed and returned from the function.

In this example you gave: The programmer is at least 'informed' in some way that the type of y is widened from number to number | string in the body of the function:

var x: { y: number } = { y: 5 };
function f(a: { y: number | string }) {
    a.y = "foo";
}

However here there is no mark of that, most users would expect full type safety as the type T should be a 'black box' to the compiler (moreover, it is returned with the unsafe assumption that it still is T, but it isn't):

class Box<T> {
    item: T;
}

function f<T extends Box<any>> (box: T) {
    box.item = "ABCD";
    return box;
}

let x = f(new Box<number>());

[Edit: removed incorrect statements, see next comment]

[malibuzios]

@Arnavion

You were right, this is not limited just to any, this was my mistake. Your first example worked better here:

class Animal {
    name: string;
}

class Dog extends Animal {
    bark() {};
}

class Cat extends Animal {
    meow() {};
}

class Box<V> {
    item: V;
}

function f<T extends Box<Animal>>(box: T) {
    box.item = new Cat(); // <-- This is perfectly valid and allowed by the constraint
    return box; // <-- The returned type is T
}

let boxWithDog = f(new Box<Dog>()); // Type of 'boxWithDog' is inferred as Box<Dog>
boxWithDog.item.bark(); // <-- Runtime error..

So the particular issue I'm talking about is the fact that f returns type T and not Box<Animal>, and that gives the 'illusion' of type safety.

(I've re-written the original comment to use this example instead)

[T18970237136]

Hi,

So the particular issue I'm talking about is the fact that f returns type T and not Box, and that gives the 'illusion' of type safety.

f returns T because it returns a value of type T and you did not specify a explicit return type. If you want it to return Box<Animal>, you can declare it as such:

function f<T extends Box<Animal>>(box: T): Box<Animal> {
    box.item = new Cat();
    return box;
}

However, the fact that you can assign a Box<Cat> to a Box<Animal> is due to TS handling generic types as covariant, as @Arnavion already said, which (I think) has to do with function arguments being considered bivariant in Typescript, whereas they are considered contravariant in C#:

let fCat: (d: Cat) => void = d => d.meow();
let fAnimal: (d: Animal) => void;

fAnimal = fCat;
fAnimal(new Dog()); // TypeError: d.meow is not a function

This is explained in the FAQ: Why are function parameters bivariant?

Edit:
For example, with the Box<V> you could think of it having both a getter and setter for item:

class Box<V> {
    private item: V;

    getItem(): V {
        return this.item;
    }
    setItem(item: V): void {
        this.item = item;
    }
}

Now, If you assign a Box<Cat> to a Box<Animal>, you assign both getItem(): Cat to getItem(): Animal, which is OK because return types are covariant (this would also work in C# if you declare the V parameter to be covariant). This means: A function that returns a Cat is also returning an Animal.

You are also assigning a setItem(item: Cat): void to setItem(item: Animal): void, which is OK in Typescript due to the function parameters being bivariant. That case however would not work in C#, as there function parameters are contravariant. In C# this means: A function which "accepts" a Cat may not accepting an Animal, e.g. a dog.

[malibuzios]

@T18970237136

What I basically say is that the fact that TypeScript allows implicit covariant casts still does not necessarily justify accepting these when deciding on compatibility with generic constraints.

You suggested the programmer should 'manually' declare the return type to improve type-safety, but isn't the whole idea of the compiler is that it should enforce type safety, well, automatically?..

[T18970237136]

Hi @malibuzios,

What I basically say is that the fact that TypeScript allows implicit covariant casts still does not necessarily justify accepting these when deciding on compatibility with generic constraints.

I'm not sure I understand you here, but I have edited my last post adding an example with getters/setters. I think you are looking for the C# behavior here, which wouldn't allow a setItem(T item) method to be declared in a Interface Box<out T> which marks T as covariant, but would allow a T getItem() method to be declared:

    interface IBox<out T>
    {
        T GetItem(); // OK
        void SetItem(T item); // Error: T is covariant but needs to be contravariant
    }

Declaring T as covariant, C# would allow you to assign a IBox<Cat> to a IBox<Animal> just like your example with a method like

    T f<T>(T box) where T : IBox<Animal>
    {
        box.SetItem(new Cat());
        return box;
    }

which you could call as f<IBox<Dog>> and get back a IBox<Dog>, but you couldn't declare a setter in IBox that accepts an item of type T.

To be honest, I expected the same behavior in Typescript and was surprised that it treats function parameter types as bivariant instead of contravariant, but I think others can explain the reasons for this much better than me.

[malibuzios]

@T18970237136

I will need some more time to fully understand what you are trying to say. But I will try to clarify because I'm not sure you have understood exactly what I meant:

1. When the T extends Box<Animal> constraint is evaluated, Box<Dog> is implicitly cast to Box<Animal> which satisfies the constraint. That would be the covariant cast.
2. When a value having type T is returned from the function, it is implicitly cast back from Box<Animal> to Box<Dog>. That would be an implicit contravariant cast!

Do you think this provides a reasonable level of type-safety, or perhaps the compiler could have done better?

[T18970237136]

Hi @malibuzios,

When a value having type T is returned from the function, it is implicitly cast back from Box<Animal> to Box<Dog>. That would be an implicit contravariant cast!

I'm not sure what you mean with this. A Box<Animal> cannot be implicitly cast to Box<Dog>:

let o1: Box<Animal> = new Box<Animal>();
let o2: Box<Dog> = o1;  // Error:
// Type 'Box<Animal>' is not assignable to type 'Box<Dog>'.
//    Type 'Animal' is not assignable to type 'Dog'.
//        Property 'bark' is missing in type 'Animal'.

Instead, you have declared f as generic function with a type parameter T that has to satisfy Box<Animal>, which Box<Dog> does, due to the generic parameter T of Box<T> being covariant. And because f returns its parameter which is of type T, f<Box<Dog>> is of type (box: Box<Dog>) => Box<Dog>.

You could also simply declare f to be non-generic and you would get the expected compile-time error:

function f(box: Box<Animal>) {
    box.item = new Cat();
    return box;
}
let boxWithDog = f(new Box<Dog>());
boxWithDog.item.bark(); // Compiler Error: Property 'bark' does not exist on type 'Animal'.

Comparing with C# (I'm using C# examples because you mentioned it in your first post), you can declare a generic function f just like in your typescript example and a Interface IBox<out T> with a covariant type parameter T:

interface IBox<out T>
{
    T Item { get; }
    // Cannot declare a Item setter with type T
}
class Box<T> : IBox<T>
{
    public T Item { get; }
    public Box(T item)
    {
        this.Item = item;
    }
}

class Animal
{
    public string name;
}
class Dog : Animal
{
    public void bark() { }
}
class Cat : Animal
{
    public void meow() { }
}

class Program
{
    static T f<T>(T box) where T : IBox<Animal>
    {
        // Do something with the animal
        Console.WriteLine("Hi, " + box.Item.name);
        // Cannot call a Setter of box.Item with type T here
        return box;
    }

    static void Main()
    {
        IBox<Dog> dog = f(new Box<Dog>(new Dog()));
        dog.Item.bark(); // OK
    }
}

The difference between C# and Typescript here is that in C# you cannot declare a setter for IBox.Item, or a method like void SetItem(T item); because T has been declared as covariant whereas for that method it would need to be contravariant (and it cannot be both); or otherwise T would need to be invariant (which would prohibit calling f<IBox<Dog>> because IBox<Dog> cannot be cast implicitly to IBox<Animal> in that case).

In TypeScript however, such a method is possible due to the function parameters being bivariant (and if you have a property item: T; you can imagine it as a replacement for a getter() and setter() method). And due to this function parameter bivariance, Generic parameters of classes and interfaces are implicitely covariant instead of invariant (I think).

But as said, others can probably explain the reasons for function parameters being bivariant better than me (as I'm relatively new to TypeScript).

[malibuzios]

@T18970237136

I will try to explain through the original example, this is really simple:

function f<T extends Box<Animal>>(box: T) {
    box.item = new Cat();
    return box;
}

let result = f(new Box<Dog>());

1. In the body of the function f, the type T is functionally equivalent to Box<Animal>. For all purposes, everything that can be done with Box<Animal>, can be done with T.
2. Now notice that Box<Animal> is a "more capable" type than Box<Dog>.
3. Now that same T type is returned back from the function (return box).
4. Now result is inferred a type. That type is Box<Dog>, not Box<Animal>.

The compiler took a type that was functionally equivalent to Box<Animal>, which is "more capable" than Box<Dog>, and cast it to Box<Dog>! This is an implicit contravariant cast.

Edit: I had a mistake in the code: I wrote new Dog() instead of new Box<Dog>(). Fixed it.