What is the reason that a function of type A => B cannot accept x => x as its input?

Question

What is the reason that a function of type A => B cannot accept x => x as its input?

It's perplexing to me why TypeScript doesn't allow the following:

type aToB = <A, B> (a:A) => B;
const id : aToB = x => x;

I keep getting the error: Type A is not assignable to type B.

What confuses me is that the type definition

type aToB = <A, B> (a:A) => B

does not imply that A must be different from B.

In mathematics, a function signature like f: A -> B does not require A to be distinct from B. For example, a function such as f: Nat -> Nat fits perfectly within f: A -> B.

Is it just TypeScript behaving oddly in this sense, or am I overlooking something obvious?

typescript

Answer 1

Answer №1

When you create a generic function, the caller has the power to determine the specific types involved. For example:

type aToB = <A, B> (a:A) => B;
declare const id: aToB = x => x;
id<number, string>(0)

In this instance, id is invoked with number and string, which aligns with the signature specified by aToB. However, if your actual implementation of id does not return a string as expected by the caller, TypeScript flags it as an error.

To address this issue, you can make aToB itself generic so that the types are determined at the point of assignment:

type aToB<A, B> =  (a:A) => B;
const id: aToB<number, number> = x => x; // This is now acceptable
id(0) // Generic types are already established
const id2: aToB<number, string> = x => x; // This would still trigger an error

Answer 2

When you create a generic function, the caller has the power to determine the specific types involved. For example:

type aToB = <A, B> (a:A) => B;
declare const id: aToB = x => x;
id<number, string>(0)

In this instance, id is invoked with number and string, which aligns with the signature specified by aToB. However, if your actual implementation of id does not return a string as expected by the caller, TypeScript flags it as an error.

To address this issue, you can make aToB itself generic so that the types are determined at the point of assignment:

type aToB<A, B> =  (a:A) => B;
const id: aToB<number, number> = x => x; // This is now acceptable
id(0) // Generic types are already established
const id2: aToB<number, string> = x => x; // This would still trigger an error

What is the reason that a function of type A => B cannot accept x => x as its input?

Answer №1

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