In the absence of a specific example, I can only provide general information. The kind of syntax you are referring to is commonly used in languages like Java that lack polymorphic this types, which I will explain briefly.
The concept revolves around defining a generic type that points to other objects of the same type as its parent class or interface. Let's take a look at your Test interface:
interface Test<T extends Test<T>> {
a: number;
b: T;
}
This sets up a structure resembling a linked list where the b property of a Test<T> must be another Test<T>, since T extends Test<T>. Moreover, it must also be (or a subtype of) the same type as the parent object. Here are two implementations to illustrate this:
interface ChocolateTest extends Test<ChocolateTest> {
flavor: "chocolate";
}
const choc = {a: 0, b: {a: 1, flavor: "chocolate"}, flavor: "chocolate"} as ChocolateTest;
choc.b.b = choc;
interface VanillaTest extends Test<VanillaTest> {
flavor: "vanilla";
}
const vani = {a: 0, b: {a: 1, flavor: "vanilla"}, flavor: "vanilla"} as VanillaTest;
vani.b.b = vani;
While both ChocolateTest and VanillaTest are implementations of Test, they are not interchangeable. The b property of a ChocolateTest is a ChocolateTest, whereas for a VanillaTest, it is a VanillaTest. Attempting to assign one to the other results in an error, thereby maintaining consistency.
choc.b = vani; // Error!
By using this approach, you can confidently navigate through the chain knowing each instance belongs to the same type:
choc.b.b.b.b.b.b.b.b.b // <-- still a ChocolateTest
Contrast this with another interface setup:
interface RelaxedTest {
a: number;
b: RelaxedTest;
}
interface RelaxedChocolateTest extends RelaxedTest {
flavor: "chocolate";
}
const relaxedChoc: RelaxedChocolateTest = choc;
interface RelaxedVanillaTest extends RelaxedTest {
flavor: "vanilla";
}
const relaxedVani: RelaxedVanillaTest = vani;
Here, RelaxedTest does not enforce the b property to be identical to the parent type but rather any implementation of RelaxedTest. Consequently, assigning instances between different flavors is permissible:
relaxedChoc.b = relaxedVani; // No error
This difference arises from the flexibility provided by RelaxedTest, allowing compatibility between various implementations irrespective of their exact types. On the contrary, with
Test<T extends Test<T>>
, such freedom is restricted based on the defined relationships.
The ability to specify a type that mandates consistency throughout the chain proves beneficial. TypeScript offers a feature known as polymorphic this designed precisely for this scenario. By utilizing this as a type representing "the same type as the containing class/interface," one can simplify the previous generic structure:
interface BetterTest {
a: number;
b: this; // <-- same as the implementing subtype
}
interface BetterChocolateTest extends BetterTest {
flavor: "chocolate";
}
const betterChoc: BetterChocolateTest = choc;
interface BetterVanillaTest extends BetterTest {
flavor: "vanilla";
}
const betterVani: BetterVanillaTest = vani;
betterChoc.b = betterVani; // Error!
This revised approach mirrors the original
Test<T extends Test<T>>
without delving into potential circular dependencies. Hence, opting for polymorphic
this presents a cleaner alternative unless there are specific reasons demanding otherwise.
If you encountered code structured differently, perhaps due to historical factors or unawareness of newer features like polymorphic this, consider this updated perspective. Good luck navigating through these concepts!
I hope this explanation clarifies the topic and aids in your understanding. Best wishes!