Why does TypeScript type inference fail in this case?

Question

I want to understand why type S1 is never, but when I remove the label or customRef properties I get correct result of string. When I remove the value and label properties I get unknown.

export interface BaseInputProps<TStored> {
  value: TStored;
  customRef?: (selfProps: this) => void;
}

export interface TestInput extends BaseInputProps<string> {
  label: string;
}

type InferStoredType<T> = T extends BaseInputProps<infer TT> ? TT : never;

type S1 = InferStoredType<TestInput>;

What is happening here?

Typescript version 3.7.5. Works the same way on the Typescript playground.

Answer 1

It has to do with the structural difference and weak type. So let's understand your problem in all the cases

Case 1: By default, it is never

When you try to extend interface TestInput with interface BaseInputProps<string> it will try to check if all the properties typing are compatible but in this case customRef?: (selfProps: this) => void; type (selfProps: this)=> void is not assignable to string and vice versa. That's why it is falsy inheritance because of which S1 is never

Case 2: When you remove label and customRef, it is string

When you removed label and customRef interface BaseInputProps and interface TestInput will be left out with one compulsory property value because of which it will correctly infer the typings.

Case 3: When you remove value and label, it is unkown

When you removed the value and label interface BaseInputProps and interface TestInput will be left out with only optional property and typescript can't guarantee the typings in this case.

Still, why this change was intentional is a question. But, looking at the scope of the spec change it will involve, I guess it's very difficult to see a change in this.

Please read this too for more on weak typing and on conditional typings.