Is there a method or alternative solution for deconstructing TypeScript types/interfaces?

Question

Is there a method or alternative solution for deconstructing TypeScript types/interfaces?

Imagine a scenario where a class has multiple type parameters:

class BaseClass<T extends T1, U extends U1, V extends V1, /* etc. */ >

Is there a method to consolidate these type arguments in a way that allows for "spreading" or "destructuring," similar to what ES6 offers for objects?

Instead of having to create subclasses like:

class FooClass extends BaseClass<TFoo, UFoo, VFoo, /* etc. */ >

Is it possible to set all type arguments at once like this:

// example of the desired functionality:
metatype FooTypes = <TFoo, UFoo, VFoo, /* etc. */>
class FooClass extends BaseClass<...FooTypes>

The purpose behind exploring this idea is that I have various objects managing different attributes of entities such as "notes" and "comments," each with multiple associated types. However, not all objects interacting with them need to handle every type.

My goal is to provide a single reference like "NoteTypes" or "CommentTypes" that essentially says "You can use any necessary type parameters from this pool."

While "type destructuring" seems like the ideal solution, the lack of support for it in TypeScript, as mentioned in this GitHub issue, poses a challenge.

In the absence of type destructuring, what would be the most effective approach to tackle this issue?

typescript

Answer 1

Answer №1

Performing object destructuring at the type level, similar to what is discussed in microsoft/TypeScript#13135, or array destructuring at the type level as mentioned in microsoft/TypeScript#5453, is not syntactically supported. However, it is possible to utilize object/tuple type indexing through lookup types. For instance, if you have an object type type M = {foo: F, bar: B} where F and B are existing types, you can retrieve F by looking up "foo" in M: M["foo"]. Similarly, B can be accessed as M["bar"].

In this context, you might consider grouping your T, U, V, etc. as a unified map of types M like so:

class BaseClass<M extends { T: T1, U: U1, V: V1 }> { /* ... */ }

To reference the T, U, and V types, you can perform lookups in the following manner:

class BaseClass<M extends { T: T1, U: U1, V: V1 }> {
    constructor(public t: M["T"], public u: M["U"], public v: M["V"]) {
    }
}

As demonstrated here, BaseClass<M> includes a constructor that accepts parameters of the original T, U, and V types.

Subclassing for convenience can be achieved as follows:

type FooTypes = { T: TFoo, U: UFoo, V: VFoo }
class FooClass extends BaseClass<FooTypes> { }

While this approach works, the process would be more straightforward if a simpler syntax for destructuring was available instead of relying on lookups. Currently, an alternative method could be to:

    type T = M["T"];
    type U = M["U"];
    type V = M["V"];

However, this method is somewhat verbose and restricted to global or function scope, not within a class (see microsoft/TypeScript#7061). In practice, utilizing M["T"] and M["U"] would be suitable wherever T and U were previously used.

Hopefully, this information proves helpful. Best of luck with your coding endeavors!

Access Playground Code Here

Answer 2

Performing object destructuring at the type level, similar to what is discussed in microsoft/TypeScript#13135, or array destructuring at the type level as mentioned in microsoft/TypeScript#5453, is not syntactically supported. However, it is possible to utilize object/tuple type indexing through lookup types. For instance, if you have an object type type M = {foo: F, bar: B} where F and B are existing types, you can retrieve F by looking up "foo" in M: M["foo"]. Similarly, B can be accessed as M["bar"].

In this context, you might consider grouping your T, U, V, etc. as a unified map of types M like so:

class BaseClass<M extends { T: T1, U: U1, V: V1 }> { /* ... */ }

To reference the T, U, and V types, you can perform lookups in the following manner:

class BaseClass<M extends { T: T1, U: U1, V: V1 }> {
    constructor(public t: M["T"], public u: M["U"], public v: M["V"]) {
    }
}

As demonstrated here, BaseClass<M> includes a constructor that accepts parameters of the original T, U, and V types.

Subclassing for convenience can be achieved as follows:

type FooTypes = { T: TFoo, U: UFoo, V: VFoo }
class FooClass extends BaseClass<FooTypes> { }

While this approach works, the process would be more straightforward if a simpler syntax for destructuring was available instead of relying on lookups. Currently, an alternative method could be to:

    type T = M["T"];
    type U = M["U"];
    type V = M["V"];

However, this method is somewhat verbose and restricted to global or function scope, not within a class (see microsoft/TypeScript#7061). In practice, utilizing M["T"] and M["U"] would be suitable wherever T and U were previously used.

Hopefully, this information proves helpful. Best of luck with your coding endeavors!

Access Playground Code Here

Is there a method or alternative solution for deconstructing TypeScript types/interfaces?

Answer №1

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