Can we guarantee the uniqueness of a function parameter value during compilation?

Question

Can we guarantee the uniqueness of a function parameter value during compilation?

I have a set of static identifiers that I want to use to tag function calls. Instead of simply passing the identifiers as arguments, I would like to ensure that each identifier is unique and throws an error if the same identifier is passed more than once:

type Id = 'id-1' | 'id-2';

function foo(id : Id) {
  console.log(`my id: ${id}`);
}

foo('id-1');
foo('id-2');

Is there a way to create a magic type in TypeScript that enforces uniqueness for the identifiers passed to the function?

type MagicId = /* some magic type to achieve uniqueness */;

function foo(id : MagicId) {
  console.log(`my id: ${id}`);
}

foo('id-1');
foo('id-1'); // should throw a type error

typescript

Answer 1

Answer №1

If you want to achieve a certain behavior, you can use objects in the following way:

type Id = 'id-1' | 'id-2';

function _foo(id: Id) {
  console.log(`My id: ${id}`)
}

function foo<T extends Id>(id : T) {
  _foo(id);
  return {
    foo(id: Exclude<Id, T>) {
      _foo(id);
    }
  }
}

foo('id-1').foo('id-2'); // this is okay

foo('id-1').foo('id-1'); // this will result in an error

The outer foo function can accept both "id-1" and "id-2", while the inner can't due to type exclusion.

This approach works best for consecutive calls, as opposed to sporadic usage of the functions.

You can try out the code on the playground.

An enhanced version could be the following:

type Builder<T> = {
  foo<I extends T>(id: I): Builder<Exclude<T, I>>;
};

const builder = {
  foo<T extends Id>(id: T): Builder<Exclude<Id, T>> {
    console.log(`My id ${id}`);
    return this;
  },
};

builder.foo("id-1").foo("id-2");

This method recursively removes the types that have already been used:

builder
  .foo("id-1") // acceptable, allows usage of "id-2"
  .foo("id-2") // all values have been used, next one defaults to `never`
  .foo("id-1") // error, as "id-1" cannot be assigned to `never`

Check out the playground for more details.

Answer 2

If you want to achieve a certain behavior, you can use objects in the following way:

type Id = 'id-1' | 'id-2';

function _foo(id: Id) {
  console.log(`My id: ${id}`)
}

function foo<T extends Id>(id : T) {
  _foo(id);
  return {
    foo(id: Exclude<Id, T>) {
      _foo(id);
    }
  }
}

foo('id-1').foo('id-2'); // this is okay

foo('id-1').foo('id-1'); // this will result in an error

The outer foo function can accept both "id-1" and "id-2", while the inner can't due to type exclusion.

This approach works best for consecutive calls, as opposed to sporadic usage of the functions.

You can try out the code on the playground.

An enhanced version could be the following:

type Builder<T> = {
  foo<I extends T>(id: I): Builder<Exclude<T, I>>;
};

const builder = {
  foo<T extends Id>(id: T): Builder<Exclude<Id, T>> {
    console.log(`My id ${id}`);
    return this;
  },
};

builder.foo("id-1").foo("id-2");

This method recursively removes the types that have already been used:

builder
  .foo("id-1") // acceptable, allows usage of "id-2"
  .foo("id-2") // all values have been used, next one defaults to `never`
  .foo("id-1") // error, as "id-1" cannot be assigned to `never`

Check out the playground for more details.

Can we guarantee the uniqueness of a function parameter value during compilation?

Answer №1

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