是否可以在 typescript 中没有明确定义的情况下创建嵌套的 generics

Question

I am having trouble writing a type that applies one type to all the bottom level props of another type.

The purpose of this is to use create a type which can convert all bottom level props to Ref<T> for Vue 3.

// Start with some type T
type T = ...

// And a 'function' to apply, U
type U<T> = ...

// This needs to apply U to all bottom level properties of T
// ('bottom level' are all props who's type does not extend object)
type DeepApply<T,U> = ???

Example:

{
  a: string,
  b: number,
  c: {
    c1: number,
    c2: string
  }
}

Would become

{
  a: Ref<string>,
  b: Ref<number>,
  c: {
    c1: Ref<number>,
    c2: Ref<string>
  }
}

Here's my best shot but TS throws an error both when trying to give a generic its own generic arguments (U<T[P]>) and when trying to pass a type with generics as a generic without explicitly defining its generics (DeepApply<T, Ref>).

type DeepApply<T, U> = {
  [P in keyof T]: T[P] extends object ? DeepUnion<T[P], U> : U<T[P]>;
};

type MyRefType = DeepApply<MyType, Ref>

Answer 1

When a type is referenced, all it's generic parameters must be satisfied. Put another way, you can't pass a parameterized type without parameters to have it's parameters filled in later. There's just no syntax for that.

However, you can do it if you hardcode the Ref type. For example:

type DeepApplyRef<T> = {
  [P in keyof T]: T[P] extends object ? DeepApplyRef<T[P]> : Ref<T[P]>;
};

I don't think a solution exists that is quite as dynamic as you want here, if your recursive type knows the type that it's wrapping things up with, then this does work just fine.

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