Typescript's Intersection Types: The Key to Overlapping Properties

Question

Typescript's Intersection Types: The Key to Overlapping Properties

Looking to create a type-safe utility function in Typescript 4.0 for comparing properties of two objects, my initial code snippet is below:

export function propertiesMatch<O extends object, T extends O, S extends O>(first: T, second: S, props: (keyof O)[]) {
    return props.every(prop => first[prop] === second[prop])
}

Encountering compile error TS2367 with this implementation, which warns that:

This condition will always return 'false' since the types 'T[keyof O]' and 'S[keyof O]' have no overlap.

This error message is puzzling as it seems logical that if both T and S extend an object of type O, they should have all keys of type O. Any insights on what I might be overlooking here and recommendations for a more robust solution would be greatly appreciated.

typescript typescript-generics

Answer 1

Answer №1

The issue you are encountering is not incorrect. Extending O simply means extending object, which can result in having T = { a: number } and O = { a: string }. The key is the same, but there is no overlap between T[keyof O] and O[keyof O]. Although there could be overlap, TypeScript expects the function to be proven correct for any valid instantiation of the type parameter, which may not always hold true.

There are different ways to define this function. One approach is to use a single type parameter for the first object and define the second object using a Pick or the passed-in properties.

function propsAreEqual<T extends object, K extends keyof T>(first: T, second: Pick<T, K>, props: K[]) {
    return props.every(prop => first[prop] === second[prop])
}

Playground Link

This version ensures that T[K] is the same type for both parameters. However, using an object literal as the second parameter may trigger excess property checks. Additionally, intellisense will suggest all properties of T in the third parameter, potentially causing errors with the second parameter if the property is not common.

I personally prefer an option that sacrifices full type checking within the function to address the aforementioned issues. Here is an alternative version:

function propsAreEqual<T extends object,  S extends Pick<T, K>,  K extends keyof T & keyof S>(first: T, second: S, props: K[]) {
    return props.every(prop => first[prop] === (second as Pick<T, K>)[prop])
}

Playground Link

Answer 2

The issue you are encountering is not incorrect. Extending O simply means extending object, which can result in having T = { a: number } and O = { a: string }. The key is the same, but there is no overlap between T[keyof O] and O[keyof O]. Although there could be overlap, TypeScript expects the function to be proven correct for any valid instantiation of the type parameter, which may not always hold true.

There are different ways to define this function. One approach is to use a single type parameter for the first object and define the second object using a Pick or the passed-in properties.

function propsAreEqual<T extends object, K extends keyof T>(first: T, second: Pick<T, K>, props: K[]) {
    return props.every(prop => first[prop] === second[prop])
}

Playground Link

This version ensures that T[K] is the same type for both parameters. However, using an object literal as the second parameter may trigger excess property checks. Additionally, intellisense will suggest all properties of T in the third parameter, potentially causing errors with the second parameter if the property is not common.

I personally prefer an option that sacrifices full type checking within the function to address the aforementioned issues. Here is an alternative version:

function propsAreEqual<T extends object,  S extends Pick<T, K>,  K extends keyof T & keyof S>(first: T, second: S, props: K[]) {
    return props.every(prop => first[prop] === (second as Pick<T, K>)[prop])
}

Playground Link

Answer 3

Answer №2

When TypeScript infers a type parameter, it tries to infer it from only one of the arguments. Therefore, the signature

propsAreEqual<T>(a: T, b: T, props: (keyof T)[])

may not work as expected because TypeScript infers T based on the first argument and does not adjust it when the second argument does not match that type.

Solution: Use a generic type that will be inferred from the props array instead of from a or b.

function propsAreEqual<K extends PropertyKey>(a: Record<K, unknown>, b: Record<K, unknown>, props: K[]) {
    return props.every(prop => a[prop] === b[prop]);
}

Playground Link

Answer 4

When TypeScript infers a type parameter, it tries to infer it from only one of the arguments. Therefore, the signature

propsAreEqual<T>(a: T, b: T, props: (keyof T)[])

may not work as expected because TypeScript infers T based on the first argument and does not adjust it when the second argument does not match that type.

Solution: Use a generic type that will be inferred from the props array instead of from a or b.

function propsAreEqual<K extends PropertyKey>(a: Record<K, unknown>, b: Record<K, unknown>, props: K[]) {
    return props.every(prop => a[prop] === b[prop]);
}

Playground Link

Typescript's Intersection Types: The Key to Overlapping Properties

Answer №1

Answer №2

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