An object does not satisfy Record<string, any> about typescript HOT 10 CLOSED

Contravariance can be confusing but that's not typescript's fault.

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Think of contravariance this way: let’s say I ask for something typed as

type Vet = (x: Animal) => Animal;

because I run a zoo and I have all sorts of animals. If what you give me instead is

type Specialist = (x: Dog | Cat) => Dog | Cat;

then that’s not going to work for me because I might need to pass in an Elephant at some point. So the subtyping relationship turns upside down when you’re the one giving me things instead of you using them yourself.

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This is working as intended. Your object does extend Record<string, any>, but (props: ModalProps) => null does not extend (props: Record<string, any>) => any. When you disable strictFunctionTypes it will work.

See also: https://github.com/microsoft/TypeScript/wiki/FAQ#why-are-function-parameters-bivariant

This is the same issue simplified:

const val = 5 satisfies number

type Example<T extends (n: number) => any> = any
type Test = Example<(n: 5) => any>

from typescript.

@MartinJohns that's a really weird and confusing behaviour. Basically I have to either remove the constraint completely or replace Record<string, any> with any

I still cannot get why (n: 5) => any does not satisfy (n: number) => any constraint, since 5 does satisfy number

I understand the intention of the strictFunctionTypes flag but here it does not seem to be the case of use

from typescript.

By the way if you make all ModalProps fields optional, then the error also goes away

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type PropTypes<TFunctionComponent> =
    TFunctionComponent extends (props: infer RProps) => any ? RProps extends Record<string, any> ? RProps : never : never;

Here is the best workaround I could have engineered for this case

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I still cannot get why (n: 5) => any does not satisfy (n: number) => any constraint, since 5 does satisfy number

The type expects a function that can accept any number, but the type passed is a function that can only accept one number (5).

Your TFunctionComponent expects a function that can accept any object (aka Record<string, any>), but you pass Modal, which is a function that expects at least ModelProps.

Your best option is to simply have PropTypes<TFunctionComponent extends (props: any) => any>.

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I get your point. The confusion comes from that I read it in the opposing order: My TFunctionComponent expects a function that can accept at least any object. ModalProps is at least any object, so everything is expected to be okay

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@fatcerberus thank you for the explication - I think I have to get used to it

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Everybody trips over this at some point; I think it naturally confuses people because it doesn’t match how we talk about things day-to-day - in everyday language both Vet and Specialist would be described as types of doctors and what we need a doctor to be able to do is something we only care about when we call one - but in a type system those requirements are encoded right into the type so you end up having to care about them sooner…

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