why does typescript function parameter type infer failed?

Question

class Base<T> {
    public state = {} as T;
    public getState(): T {
        return this.state
    }
    public setState(v: T) {
        this.state = v
    }
}

interface DogProps {
    name: 'hello';
    age: 123;
}

class Dog extends Base<DogProps> {
    public sayName() {
        console.log('name: ', this.state.name);
    }
    public sayAge() {
        console.log('age: ', this.state.age);
    }
}

function test<U, T extends Base<U>>(Cor: new () => T): [U, T] {
    const dog = new Cor();
    const state = dog.getState();
    return [state, dog];
}

const [state1, dog1] = test(Dog); // state1 is unknow

const [state2, dog2] = test<DogProps, Dog>(Dog); // verbose but right

demo playground

I am newbe in typescript.
I thought the code I wrote was right. But it does not work as expected.
Why state1's type is unknow?
Can I get the right type without test<DogProps, Dog>(Dog)?

much thanks!!!

Tadhg McDonald-Jensen · Accepted Answer · 2020-09-25 14:15:40Z

3

This is a side effect of how generic resolution works, typescript sees that T is referred to in the arguments so it tries to resolve it, but the constraint is based on U so it tries to resolve that first. Because U doesn't appear anywhere in the argument list, it can't resolve it so it ends up unknown

If you ensure that U is present in the arguments list you can ensure that typescript will be able to resolve it from just looking at the input without having to figure out T first:

function test<U, T extends Base<U>>(Cor: new()=>(T & Base<U>)): [U, T] {
                                                 // ^here^
}

This should fix the issue :)

edited Sep 25, 2020 at 14:15

answered Sep 16, 2020 at 16:11

Tadhg McDonald-Jensen

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6 Comments

threetree Over a year ago

You are my god. You saved my life.

Tadhg McDonald-Jensen Over a year ago

I sure hope your life doesn't depend on something like this 🤣. Happy coding! 😁

threetree Over a year ago

Now I understood the first paragraph, the reason why my code does not work. But I still not understand the second paragraph, why your code works. Especially without having to figure out T first. Because return param still need type T. Could you explain more details?

Tadhg McDonald-Jensen Over a year ago

pretend you are typescript, you are looking at the types of arguments and you are trying to figure out what U should resolve to. But U isn't specified anywhere in the input arguments, only T, so how can you figure out what U is? At some point typescript will probably get updated to be able to cope with this case but at least for today using (T & (Constraint of T)) works fine.

threetree Over a year ago

As a human being, we know Dog extends Base, then U's type must be DogProps. But typescript seems not smart enough. By the way, could you recommend any books or website or articles to learn advanced typescript skills, like how generic resolution works. It seems that typescript document does not have these things.

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