为什么打字稿函数参数类型推断失败? [英] why does typescript function parameter type infer failed?

问题描述

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

演示操场

我是打字稿新手.
我认为我写的代码是对的.但它没有按预期工作.
为什么 state1 的类型是未知的?
我可以在没有 test(Dog) 的情况下获得正确的类型吗?

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)?

非常感谢！！！

推荐答案

这是泛型解析工作方式的副作用，打字稿看到 T 在参数中被引用，因此它试图解析它，但约束是基于 U 所以它试图首先解决这个问题.因为 U 没有出现在参数列表中的任何地方，它无法解析它所以它以 unknown

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

如果您确保 U 出现在参数列表中，您可以确保打字稿能够通过查看输入来解决它，而无需弄清楚 T 首先:

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^
}

这应该可以解决问题:)

This should fix the issue :)

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