curryRecord

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curryRecord

This is an example of a function whose logic lives in the type definition rather than in the function body.
/**
 * @name `curryRecord`
 *
 * Takes a unary function that takes a record as an argument, and makes the
 * record partially applicable. Returns a new function that takes a partial of
 * the record, which itself returns a new function that takes only the missing
 * properties of the record.
 *
 * Can be thought of as `curry()` that curries a record as opposed to a list of
 * arguments.
 */
export const curryRecord = <F extends (x: Record<any, any>) => any>(fn: F) => <
  T extends Partial<Parameters<F>[0]>
>(
  a: T
) => (b: Omit<Parameters<F>[0], keyof T>): ReturnType<F> =>
  fn({ ...a, ...b } as Parameters<F>[0]);
Breaking it down
export const curryRecord = <F extends (x: Record<any, any>) => any>(fn: F) => ...
The first argument is a unary function. We use the generic <F extends...> to enforce arity and to enforce the passed function's argument being a record. The exact type of x and the return type will be inferred by Typescript once a function is actually passed.
<T extends Partial<Parameters<F>[0]>>(a: T) => ...
Once we pass our unary function we get a new function that takes a: T. T must be assignable to Partial<Parameters<F>[0]>, which means "a partial version of whatever record your unary function expects as an argument."
(b: Omit<Parameters<F>[0], keyof T>): ...
Once we pass our partial record, we get a new function that takes a record b that must contain all keys of Parameters<F>[0] that were NOT included in a. It checks this by omitting all keys of a from Parameters<F>[0]: the remaining keys are the keys that must be passed as a part of b.
ReturnType<F>
The return type of the curried function should match the return type of the function we're currying in the first place. We don't know what this is, so we use ReturnType<F> to infer the proper type once a function is actually passed.
... => fn({ ...a, ...b } as Parameters<F>[0])
We trivially spread both partial arguments into the original function. We have to coerce this here because Typescript can't figure out that the above constraints means that { ...a, ...b } will always be of type Parameters<F>[0].
Interestingly enough, once compiled to Javascript this function becomes curryRecord => fn => a => b => fn({ ...a, ...b }), which does absolutely zero validation. All the logic is handled by the type system!
Demo and test cases
interface FooBar {
  foo: string;
  bar: string;
}
​
const handleFooBar = (foobar: FooBar): string => foobar.foo + foobar.bar;
​
// Works trivially
handleFooBar({
  foo: "foo",
  bar: "bar"
});
​
// TypeError: Property "bar" is missing
handleFooBar({
  foo: "foo"
});
​
// Returns a version of handleFooBar ready for partialization
const curryRecordHandleFooBar = curryRecord(handleFooBar);
​
// Returns a function that only allows the missing properties
const handleFooBarWithFooProvided = curryRecordHandleFooBar({
  foo: "foo"
});
​
// TypeError: "foo" is not assignable to Omit<FooBar, 'foo'>
handleFooBarWithFooProvided({ foo: "foo2" });
​
// TypeError: Property "bar" is missing
handleFooBarWithFooProvided({});
​
// Works!
handleFooBarWithFooProvided({ bar: "bar" });
​
// There are no missing properties, hence this is called with an empty record
const handleFooBarWithAllKeysProvided = curryRecordHandleFooBar({
  foo: "foo",
  bar: "bar"
});
​
handleFooBarWithAllKeysProvided({});
​
// TypeError: rejects functions that do not take records
curryRecord((x: number) => 123);
​
// TypeError: rejects non-unary functions
curryRecord((x: Record<any, any>, y: Record<any, any>) => 123);