TypeScript is unable to infer type properly from a typed string variable

Question

I am trying to use a "freely" created string as a key for an object.

interface CatState {
    name: string,
    selected: boolean,
    color: string,
    height: number
}

interface DogState{
    name: string,
    selected: boolean,
    race: string,
    age: number,
    eyeColor: string
}

export interface Animals {
    animals: {
        cat: {
            cats: CatState[],
            allSelected: boolean,
        },
        dog: {
            dogs: DogState[],
            allSelected: boolean,
        },
    }
};

const selectAnimal = (allAnimals: Animals, animal: keyof Animals['animals'], index:number) => {
    const animalPlural = `${animal}s` as keyof Animals['animals'][typeof animal]
    allAnimals.animals[animal][animalPlural][index].selected= true
}

This highlights .selected with the message

Property 'selected' does not exist on type 'boolean'.

Here is a Playground. Is there a workaround for this, or is this simply not possible?

Answer 1

In order for this to work you need to make selectAnimal generic.

You might think it should be able to deal with an animal input of a union type, but the compiler isn't able to properly type check a single block of code that uses multiple expressions that depend on the same union type. It loses track of the correlation between `${animal}s` and allAnimals.animals[animal]. The formers is of type "cats" | "dogs" and the latter is of a type like {cats: CatState[]} | {dogs: DogState[]}, and you can't generally index into the latter with the former, because "what if you've got {cats: CatState[]} and you're indexing with "dogs"?" That can't happen, but the compiler is unable to see it. TypeScript can't directly deal with correlated unions this way. That's the subject of microsoft/TypeScript#30581.

If you want a single code block to work for multiple cases, the types need to be refactored to use generics instead, as described in microsoft/TypeScript#47109. Here's how it might look for your example:

interface AnimalStateMap {
    cat: CatState,
    dog: DogState
}

type AnimalData<K extends keyof AnimalStateMap> =
    { [P in `${K}s`]: AnimalStateMap[K][] & { allSelected: boolean } }

export interface Animals {
    animals: { [K in keyof AnimalStateMap]: AnimalData<K> };
};
    

const selectAnimal = <K extends keyof AnimalStateMap>(
    allAnimals: Animals, animal: K, index: number) => {
    const animalPlural = `${animal}s` as const;
    // const animalPlural: `${K}s`
    const animalData: AnimalData<K> = allAnimals.animals[animal]
    animalData[animalPlural][index].selected = true;
}

The AnimalStateMap is a basic key-value type representing the underlying relationship in your data structure. Then AnimalData<K> is a mapped type that encodes as a template literal type the concatenation of s onto the type of the keys (giving such plurals as gooses and fishs 🤷‍♂️) and that the value type is of the expected animal array. And that there's an allSelected property.

Then your Animals type explicitly written as a mapped type over keyof AnimalStateMap, which will help the compiler see the correlation when we index into it.

Finally, selectAnimal is generic in K extends keyof AnimalStateMap and the body type checks because animalPlural is of just the right generic type `${K}s` which is known to be a key of animalData, which is AnimalData<K>.

Playground link to code

Answer 2

You should use a similar variable name (e.g. states) inside FooState.animals[type] instead of cats and dogs. Without doing so, TypeScript sees cats and dogs as two unrelated objects.

See this TypeScript playground. Note that I've the removed PayloadAction type since the OP did not provide it and added a type to the object destructuring.

FooState can also be converted from a regular type to an interface. Also notice I've add a FooStateAnimalEntry interface. Adding this interface gives you more organized code and improves readability.

You will probably want to change the type of FooStateAnimalEntry.more, which may need to be done with generics.