类类型的字典作为参数，其实例的字典作为返回类型

Question

我想编写一个函数，该函数接受值是类的字典作为参数，并且在输出中我想获取包含其实例的字典

例如

classes C1, C2
function f: ({
    name1: C1,
    name2: C2
}): ({
    name1: new C1()
    name2: new C2()
})

如何为函数f编写类型以获得强类型？

Answer 1

是的，您可以使用映射类型和内置条件类型InstanceType

class C1 { c1!: number} 
class C2{ c2!: number }
type AllInstance<T extends Record<string, new (...a: any[]) => any>> = {
    [P in keyof T]: InstanceType<T[P]>
}
function f<T extends Record<string, new () => any>>(o: T): AllInstance<T> {
    const result = {} as AllInstance<T>
    for(var key of Object.keys(o) as Array<keyof T>) {
        result[key] = new o[key]()
    }
    return result;
}

let o = f({ C1,C2 })
o.C1.c1;
o.C2.c2;

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