如何推断泛型函数的返回类型

问题描述：

我有一个总是返回一个类型的函数，虽然它可以改变并手动定义它会是一些工作并且不可扩展，所以我试图通过使用打字稿的 infer 来实现这一点关键字

I have a function that will always return a type, though it can change and defining it by hand would be some work and not scalable, so I'm trying to achieve this by using typescript's infer keyword

起初我看到了这个reddit帖子使用

type Foo<T> = T extends { a: infer U, b: infer U } ? U : never;
type T10 = Foo<{ a: string, b: string }>;  // string
type T11 = Foo<{ a: string, b: number }>;  // string | number

问题是，我的类型是一个泛型函数

Problem is, my type is a generic function

这是一个最小的例子，

const fnFromLib = <T>(a: string, b: Record<string, string>) => {
  return function barFn(c: T) {
    return {
      c,
      b: b[a],
    };
  };
};
const foo = <T>(a: Record<string, string>) => {
  return {
    bar: fnFromLib<T>('foo', a),
  };
};
type returnTypeInferer<T> = T extends (a: Record<string, string>) => infer U ? U : never;
type fooType = typeof foo;
type fooReturnType = returnTypeInferer<fooType>;

没有错误，但是，由于没有传递泛型类型，fooReturnType 将被推断为

there are no errors, but, since there was not a generic type being passed, fooReturnType will be inferred as

type fooReturnType = {
    bar: (c: unknown) => {
        c: unknown;
        b: string;
    };
}

注意到 fnFromLib 中的 T 不是从参数推断的，应该在函数调用中传递我尝试将类型参数从 fooReturnType 传递到 fooType，但是给我一个解析错误

Noting that T from fnFromLib is not inferred from arguments and should be passed in the function call I try to pass on a type argument from fooReturnType to fooType, but that gives me a parsing error

type returnTypeInferer<T> = T extends (a: Record<string, string>) => infer U ? U : never;
type fooType<T> = typeof foo<T>; // ------   parsing error: ';' expected ---------
type fooReturnType<T> = returnTypeInferer<fooType<T>>;

有什么方法可以实现我想要的吗?

Is there a way I can achieve what I want?

谢谢

答

通过将函数包装到泛型类中来设法做到这一点

Managed to do it by wrapping the function into a generic class

class GnClass<T> {
  foo = (a: Record<string, string>) => {
    return {
      bar: fnFromLib<T>('foo', a),
    };
  };
}
type returnTypeInferer<T> = T extends (a: Record<string, string>) => infer U ? U : never;
type fooType<T> = GnClass<T>['foo'];
type fooReturnType<T> = returnTypeInferer<fooType<T>>;

但我想坚持使用函数式编程，我不会将此标记为已接受的答案，暂时将使用它，因为我目前看不到其他替代方法，并且会喜欢另一种方法.

But I want to stick with functional programming, I will not mark this as accepted answer, will use it for now as I see no other alternative in the moment and would love another approach.

如何推断泛型函数的返回类型

相关推荐