Universal Parameter Typing in Functions

I'm grappling with a concept that seems obvious to me, yet is disallowed by Typescript when all strict flags are enabled (presumably for valid reasons). Let me illustrate:

We all understand the following:

export interface Basic {
  value: "foo" | "bar" | "baz";
}

export interface Constraint extends Basic {
  value: "foo";
}

It works as expected. However, something more surprising (at least to me) is:

export interface WithFunctionsBasic {
  value: (t: "foo" | "bar" | "baz") => void;
}

export interface WithFunctionsConstraint extends WithFunctionsBasic { // Incorrectly extends WithFunctionsBasic
  value: (t: "foo") => void;
}

This does not work.

We can introduce generics to achieve this kind of behavior like so:

export interface WithFunctionsParametric<T extends Basic> {
  value: (t: T["value""]) => void;
}

const variableConstraint: WithFunctionsParametric<Constraint> = { value: (t: "foo") => {} };
const variableBasic: WithFunctionsParametric<Basic> = { value: (t: "foo" | "bar" | "baz") => {} };

Both of these examples work. But it becomes slightly complicated because despite Constraint extending Basic:

const variableConstraint: WithFunctionsParametric<Constraint> = { value: (t: "foo") => {} };
const variableBasic: WithFunctionsParametric<Basic> = variableConstraint; // Does not work

In my understanding, if Constraint extends Basic (which it does), the latter should work. Can someone please explain why this isn't the case?

Answer №1

Your core issue arises from a key distinction between the behavior of inputs and outputs in terms of inheritance.

Let's consider the initial example:

export interface Base {
  value: "foo" | "bar" | "baz";
}

export interface Constraint extends Base {
  value: "foo";
}

In this scenario, Constraint extends Base. Both interfaces have the same property value, but they differ in its type definition.

  • Base specifies that value can be either "foo", "bar", or "baz".
  • Constraint narrows it down to only "foo", which is acceptable as it falls within the valid values defined by Base.

In essence, an instance of Constraint does not violate the constraints set by Base. The property value will always belong to the set "foo", "bar", or "baz". Even though Constraint restricts it further, it still upholds the promise that the value belongs to one of those three options.

Now moving on to the second example:

export interface WithFunctionsBase {
  value: (t: "foo" | "bar" | "baz") => void;
}

export interface WithFunctionsConstraint extends WithFunctionsBase { // WithFunctionsConstraint incorrectly extends WithFunctionsBase
  value: (t: "foo") => void;
}

What does this signify? Well, WithFunctionsBase guarantees: "I possess a function named value that accepts either a "foo", a "bar", or a "baz"".

However, WithFunctionsConstraint extending WithFunctionsBase fails to fulfill this assurance. Its value function now only takes a "foo". Essentially, we are attempting to define a subtype of

WithFunctionsBase</code that violates the contract established by the parent interface since its <code>value
function no longer accommodates "bar" or "baz".

Regarding outputs, a subtype must maintain the same or stricter restrictions. Narrowing down the output types doesn't breach the commitments made by the supertype. In formal terms, return types exhibit covariance; they can become more specialized through inheritance.

Conversely, concerning inputs, a subtype should maintain the same or looser restrictions. It must at least accept the same inputs (to uphold the agreement), while also having the option to allow additional types not permitted by the supertype. Function arguments adhere to contravariance; they can become less specific via inheritance.

The current setup with WithFunctionsConstraint solely accepting "foo" is invalid because it means that a WithFunctionsConstraint does not fully align with being a valid WithFunctionsBase. However, allowing it to accept "foo", "bar", "baz", "lol", and "wut" would be entirely valid.

The other challenges you're encountering in the query all appear to stem from variations of this fundamental problem.

Answer №2

When it comes to extending, the key lies in what you choose to extend. In the initial example, the interface Basic is defined with a value property that can have values of "foo", "bar", or "baz".
On the other hand, when you try to override this with Constraint, it still aligns with the interface because "foo" is included in the list of possible values.

Now, looking at the second example, the statement

value: (t: "foo" | "bar" | "baz") => void;
operates differently than expected. This defines a function with one parameter out of the 3 options. Consequently, the extended class no longer matches this criteria since the function definitions do not align. For instance, value: (t: "foo") => void; is not encompassed within
value: (t: "foo" | "bar" | "baz") => void;

If your intention is indeed to segment as proposed, you would need to split the basic function into distinct calls with individual types like so:

export interface WithFunctionsBasic {
  value: ((t: "foo") => void) | ((t: "bar") => void) | ((t: "baz") => void);
}

export interface WithFunctionsConstraint extends WithFunctionsBasic {
  value: (t: "foo") => void;
}

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