Ensuring type integrity for intersections containing varying numbers of elements

Currently, I am navigating a sophisticated custom typeguard library developed for a project I'm involved in. I am facing challenges in grasping the concept of function signatures used in typeguards.

The library includes a generic Is function that has this structure:

type Is<A> = (a: unknown) => a is A

This setup allows me to create composable typeguards like:

const isString: Is<string> = (u: unknown): u is string => typeof u === 'string'
const isNumber: Is<number> = (u: unknown): u is number => typeof u === 'number'

There are also similar implementations for records, structs, arrays, and more. For instance, the array typeguard looks like this:

const isArray = <A>(isa: Is<A>) => (u: unknown): u is A[] => Array.isArray(u) && u.every(isa)

Then there's one specifically for objects:

export const isStruct = <O extends { [key: string]: unknown }>(isas: { [K in keyof O]: Is<O[K]> }): Is<O> => (
  o
): o is O => {
  if (o === null || typeof o !== 'object') return false
  const a = o as any
  for (const k of Object.getOwnPropertyNames(isas)) {
    if (!isas[k](a[k])){
      return false
    }
  }
  return true
}

An example usage would be:

const isFoo: Is<{foo: string}> = isStruct({foo: isString})

We have a simplistic overloaded isIntersection function currently in place:

export function isIntersection<A, B>(isA: Is<A>, isB: Is<B>): (u: unknown) => u is A & B
export function isIntersection<A, B, C>(isA: Is<A>, isB: Is<B>, isC: Is<C>): (u: unknown) => u is A & B & C
export function isIntersection<A, B, C>(isA: Is<A>, isB: Is<B>, isC?: Is<C>) {
  return (u: unknown): u is A & B & C => isA(u) && isB(u) && (!isC || isC(u))
}

The issue arises when adding additional typeguards beyond three since nesting isIntersection functions becomes necessary.

Building on insights from @jcalz, particularly his responses in the following link: Typescript recurrent type intersection, I came up with the following Intersection type:

type Intersection<A extends readonly any[]> =
  A[number] extends infer U ?
    (U extends any ? (k: U) => void : never) extends ((k: infer I) => void) ?
      I : never : never;

Based on this type, here is a potential implementation for the guard:

export function isIntersection<T extends any[]>(...args: { [I in keyof T]: Is<T[I]> }): Is<Intersection<T>>
{
    return (u: unknown): u is Intersection<T[number]> => args.every((isX) => isX(u))
}

Although this solution works, the mystery lies in how the Intersection type correctly infers the specific type.

I extend my appreciation to @jcalz for providing valuable answers and encouraging clarity in understanding these concepts.

Playground

Answer №1

My proposed approach in this scenario is to define the generic type parameter T in the isIntersection function to represent a tuple of the types being guarded for by Is<T>. For example, if you have isA as Is<A> and isB as Is<B>, then calling isIntersection(isA, isB) should be generic with regard to the type [A, B]. By leveraging mapped tuple types, we can specify both the input arguments (args) and the return type in terms of T.

For instance:

function isIntersection<T extends any[]>(
  ...args: { [I in keyof T]: Is<T[I]> }
): Is<IntersectTupleElements<T>> {
  return (u: unknown): u is IntersectTupleElements<T> =>
    args.every((isX) => isX(u))
}

The args list represents a mapped type where each element of T is encapsulated with Is<>. Therefore, if T is [A, B], then args will be of type [Is<A>, Is<B>]. The resulting type is

Is<IntersectTupleElements<T>>
, where IntersectTupleElements<T> derives the intersection of elements from the tuple, like A & B.

We can implement it like this:

type IntersectTupleElements<T extends any[]> =
  { [I in keyof T]: (x: T[I]) => void }[number] extends
  (x: infer I) => void ? I : never;

This method resembles the process used in UnionToIntersection<T> discussed in this question/answer, but specifically operates on tuple elements instead of union members. When dealing with a type like [A | B, C], the goal for IntersectTupleElements<T> is to yield (A | B) & C. Ensuring that [A | B, C] isn't conflated into (A | B | C)[] results in A & B & C which may not be desired.


To test out our implementation, consider the following:

interface A { a: string }
interface B { b: number }
const isA: Is<A> = isStruct({ a: isString });
const isB: Is<B> = isStruct({ b: isNumber });

const isAB = isIntersection(isA, isB)
// function isIntersection<[A, B]>(args_0: Is<A>, args_1: Is<B>): Is<A & B>
// const isAB: Is<A & B>

The output shows that invoking isIntersection(isA, isB) prompts the compiler to infer T as [A, B], leading to a return type of Is<A & B>. Moreover, as this is variadic, utilizing

isIntersection(isA, isB, isC, isD, isE
would generate a result of type
Is<A & B & C & D & E>
, and so forth.

Check out the code in the Playground

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