Why does the inferred type not get resolved in the else block of the ternary operator when using infer on a generic type?

Question

Why does the inferred type not get resolved in the else block of the ternary operator when using infer on a generic type?

Consider a scenario where we have a type with a generic argument T:

type Wrap<T> = { data: T };

If we define another type to extract the generic T:

type Unwrap<W> = W extends Wrap<infer T> ? T : T;

Why does using T in the else clause result in the error: Cannot find name 'T'?

I am aware that using never resolves this issue, but in practice, I require something more complex involving the second T.

Check out the Stackblitz example for further clarification: https://stackblitz.com/edit/angular-k7axek?file=src%2Fmain.ts,src%2Fform-model.ts

typescript typescript-generics

Answer 1

Answer №1

One issue may arise due to the fact that T would not have anything to reference in the false branch of the conditional statement. In a scenario where Wrap<infer T> is false, T cannot be inferred, and even if it did exist, it would lack significance. Therefore, it is only declared within the true branch and not in the false one.

The official documentation does not explicitly mention this concept. However, it does indicate that the inferred type is declared within the true branch. The release notes for version 2.8 (when infer was first introduced) provide more direct insight into this:

In the extends clause of a conditional type, you can now use infer declarations to introduce a type variable that needs inference. This type variable must be referenced in the true branch of the conditional type.

(emphasis added)

Although the explanation remains vague, my interpretation above is based on inferred reasoning, which I believe to be valid.

Answer 2

One issue may arise due to the fact that T would not have anything to reference in the false branch of the conditional statement. In a scenario where Wrap<infer T> is false, T cannot be inferred, and even if it did exist, it would lack significance. Therefore, it is only declared within the true branch and not in the false one.

The official documentation does not explicitly mention this concept. However, it does indicate that the inferred type is declared within the true branch. The release notes for version 2.8 (when infer was first introduced) provide more direct insight into this:

In the extends clause of a conditional type, you can now use infer declarations to introduce a type variable that needs inference. This type variable must be referenced in the true branch of the conditional type.

(emphasis added)

Although the explanation remains vague, my interpretation above is based on inferred reasoning, which I believe to be valid.

Answer 3

Answer №2

If a class doesn't inherit from Wrap<T>, it can take on various forms such as a number, string, array, or another object. In this scenario, the actual type remains ambiguous and unknown to you.

When the possibilities are limited - for example, if W can only be a Wrap or a string - you can use nested extends like this:

type Unwrap<W> = W extends Wrap<infer T> ? T : W extends string ? string : unknown;

Upon revisiting your question, it seems like what you're looking for is this:

type Unwrap<W> = W extends Wrap<infer T> ? T : W;

Answer 4

If a class doesn't inherit from Wrap<T>, it can take on various forms such as a number, string, array, or another object. In this scenario, the actual type remains ambiguous and unknown to you.

When the possibilities are limited - for example, if W can only be a Wrap or a string - you can use nested extends like this:

type Unwrap<W> = W extends Wrap<infer T> ? T : W extends string ? string : unknown;

Upon revisiting your question, it seems like what you're looking for is this:

type Unwrap<W> = W extends Wrap<infer T> ? T : W;

Why does the inferred type not get resolved in the else block of the ternary operator when using infer on a generic type?

Answer №1

Answer №2

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