typescript - TypeScript 提供（未指定）泛型类型作为泛型参数

Question

我希望提供一个泛型类型作为类型参数，而不首先将其解析为具体类型。换句话说，我正在寻找一种方法来指定从基类继承时可以使用的类型映射函数。

示例（不正确）语法，希望比我能解释得更好：

abstract class RunIt<SomeMagicConverter> {
    // The return type for this function depends on the type of the
    // argument *and* on the way the implementation class was declared:
    run<T>(o: T): SomeMagicConverter<T> { // (syntax error)
        return this.handle(o); // (imagine this does something more interesting)
    }
    protected abstract handle<T>(o: T): SomeMagicConverter<T>; // (syntax error)
}

type MyMagicConverter<T> = TypeIWantToReturn<T>; // MyMagicConverter is generic
class MyRunIt extends RunIt<MyMagicConverter> { // but we don't specify T here
    // [...]
}
new MyRunIt().run(7); // call infers T is number, so returns TypeIWantToReturn<number>
new MyRunIt().run(''); // now T is string, so returns TypeIWantToReturn<string>

此外，我想限制这一点，以便SomeMagicConverter<T> extends SomeBase<T>得到保证。IE

abstract class RunIt<SomeMagicConverter extends SomeBase>

---

对于我希望如何使用它的更具体的示例，这是一个带有缓存的包装的基本基类（不是我的实际用例，但演示了需要）：

interface Wrapped<T> {
    contains(other: T): boolean;
}

abstract class Store {
    private readonly cached = new Map<any, any>();

    protected abstract applyWrap<T>(o: T): Wrapped<T>;

    wrap<T>(o: T): Wrapped<T> { // <-- this should return something more specific
        if (!this.cached.has(o)) {
            this.cached.set(o, this.applyWrap(o));
        }
        return this.cached.get(o);
    }
}

class Foo<T> implements Wrapped<T> {
    constructor(private readonly o: T) {}
    contains(other: T): boolean { return other === this.o; }
    extraFooFunc(): void {}
}

class FooWrapper extends Store {
    constructor() { super(); }
    protected applyWrap<T>(o: T): Foo<T> { return new Foo<T>(o); }
}

new FooWrapper().wrap(4).extraFooFunc(); // syntax error because extraFooFunc is not defined on Wrapped

显然我可以通过定义包装方法来解决这个问题，但我想避免在每个子类上都这样做：

class FooWrapper extends Store {
    // [...]
    wrap<T>(o: T): Foo<T> { return super.wrap(o) as Foo<T>; }
}

标签： typescripttypescript-generics

Answer 1

尚不支持泛型泛型参数（2021 年 3 月）。有关更多信息，请参阅TypeScript 问题和类似的 SO 问题。

但是，您所描述的可以使用Indexed Access Types和Mapped Types来实现：

// Any Converter to be used with RunIt must be added to Converters
interface Converters<T> { }

// All Converters must implement this interface
interface ConverterBase<T> { length: number }

// Maps valid keys to themselves, invalid ones to `never`
type ValidKey<T, K extends keyof Converters<T>> =
    Converters<T>[K] extends ConverterBase<T> ? K : never;

// Contains all entries from Converters where the type extends ConverterBase
type ConstrainedConverters<T> = {
    [K in keyof Converters<T> as ValidKey<T, K>]: Converters<T>[K];
};

abstract class RunIt<K extends keyof ConstrainedConverters<void>> {
    run<T>(o: T): ConstrainedConverters<T>[K] {
        return this.handle(o);
    }
    protected abstract handle<T>(o: T): ConstrainedConverters<T>[K];
}

// Re-open Converters and add some Converters
interface Converters<T> { Id: T, Array: Array<T> }

class ArrayRunIt extends RunIt<'Array'> {
    protected handle<T>(o: T) { return [o]; }
}

// @ts-expect-error Type '"Id"' does not satisfy the constraint '"Array"'.(2344)
class IdRunIt extends RunIt<'Id'> {
    protected handle<T>(o: T) { return o; }
}