Determining the generic type argument of a class can be unsuccessful due to the specific properties within that class

Why is it that Typescript sometimes fails to infer types in seemingly simple cases? I am trying to understand the behavior behind this.

When Typescript's Type Inference Goes Wrong

Consider the scenario where we have the following class declarations:

declare class Swine
{
    grunt(): void
}

declare abstract class Part<T> {}

declare class Hoof extends Part<Swine> {}

And a function like this:

declare function getOwner<T>(part: Part<T>): T;

Now, if we try to execute the following code snippet:

getOwner(new Hoof())
  .grunt();

We'll find that the Typescript compiler throws an error because it infers the return type of getOwner as unknown instead of Swine.

The Unconventional Solution that Works

Surprisingly, a simple addition of a property animal: T to the declaration of Part resolves the issue and makes the code functional.

In other words, updating the definition of Part to include:

declare abstract class Part<T>
{
  animal: T
}

Enables Typescript to correctly infer the return type of getOwner(new Hoof()). This behavior has been confirmed for Typescript versions 3.9.7 and 4.1.3.

An Inquiry into Typescript's Behavior

Given this peculiar workaround, one must question why Typescript acts in this manner. Shouldn't it retain generic type information even when T is not directly used in any class members' signatures?

Answer №1

To find more information on unused type parameters in TypeScript, you can refer to the TypeScript FAQ section.

The type system of TypeScript is primarily structural rather than nominal. This means that when comparing two types, such as A and B, they are considered the same if their structures match (e.g., both are object types with members of the same keys and value types). The names and declarations of the types do not affect this comparison.


For instance, consider the abstract class declaration:

declare abstract class Part<T> { }

In this case, regardless of the type specified for T, the resulting type is an empty object type with no members, making it equivalent to just {}. Therefore, Part<swine> is indistinguishable from Part<string> or any other type.

Adding a property of type T to Part<T> will differentiate between types like Part<Swine> and Part<string>, allowing the compiler to infer T correctly.


At the value level, similar behavior can be seen by examining functions that manipulate string representations of types. If the output does not depend on the input, inference becomes impossible. On the other hand, if the function output varies based on the input, inference becomes feasible.


Explore this concept further using the provided Playground link.

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