[SOLVED] Why is the infer keyword needed in Typescript?

Why is the infer keyword needed in Typescript?

Why did the Typescript folks create the infer keyword? According to the documents, this is an example of how you would use it:

type ReturnType<T> = T extends (...args: any[]) => infer R ? R : any;

I don't understand why this is needed. Why can't it just be:

type ReturnType<T> = T extends (...args: any[]) => R ? R : any;

Why doesn't this work? Why is the infer keyword necessary ?

Solution

With infer, the compiler ensures that you have declared all type variables explicitly:

type MyType<T> = T extends infer R ? R : never;
type T1 = MyType<{b: string}> // T1 is { b: string; }

Here we declare a new type variable R in MyType, which gets inferred from T.
Note that infer is only used within the extends clause of a conditional type.

Without infer, the compiler does not introduce an additional type variable R2 that is to be inferred (see first case). If R2 has not been declared, it will result in a compile error:

type MyType2<T> = T extends R2 ? R2 : never; // error, R2 undeclared

More precisely, the compiler checks if T is assignable to R when infer is omitted:

type R = { a: number }
type MyType3<T> = T extends R ? R : never; // compare T with type R
type T3 = MyType3<{b: string}> // T3 is never

Note, that infer R shadows type references of an equally-named type declaration R:

type R = { a: number }
type MyType4<T> = T extends infer R ? R : never;
type T4 = MyType4<{b: string}> // { b: string; }

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