请教一个类型推断的小问题，如何省去不需要传的参数返回正确的值？

Question

演示地址：https://tsplay.dev/N9De8w

要求：根据提供的参数推断返回的类型（已实现）
问题：

• 第二个参数类型是：Fn<D, any> | undefined
• 我本意是可以不传这个参数
• 但不传这个参数，拿到的结果就是 any

怎么才能让第二个参数不传也能得到结果

示例代码：

function tableSubmit <
  T extends FormInstance<any>,
  D extends DataFormType<T>,
  F extends Fn<D, any> | undefined,
  P extends FilterFormat<F, D>
>(
  form: T, filter?: F
): P {
  const params: D = form.get();
  const data  = filter !== undefined ? filter(params) : params;
  return data;
}

interface FormInstance<T> {
  data: T
  get: () => T;
}

type Fn<D, T extends any> = (data: D) => T 

type DataFormType<T> = T extends FormInstance<infer U> ? U : T

type FilterFormat<F, D> = F extends (data: D) => infer P ? P : D

const nameData = { name: "levi" };
const formName: FormInstance<{ name: string }> = {
  data: nameData,
  get: () => nameData
};

const ageData = { age: 18 };
const formAge: FormInstance<{ age: number }> = {
  data: ageData,
  get: () => ageData
};


const dataName = tableSubmit(formName, data => ({ ...data, site: 'sh' }));
const dataAge = tableSubmit(formAge, undefined);

// remove undenfind args is any
const dataError = tableSubmit(formAge);

Answer 1

知道怎么做了

（推荐）方法 1：重载 https://tsplay.dev/w6D0Em

function tableSubmit <T, R>(form: FormInstance<T>): T 
function tableSubmit <T, R>(form: FormInstance<T>, filter: undefined): T 
function tableSubmit <T, R>(form: FormInstance<T>, filter: (data: T) => R): R
function tableSubmit <T, R>(form: FormInstance<T>, filter?: (data: T) => R): R | T {
  const params = form.get();
  const data  = filter !== undefined ? filter(params) : params;
  return data;
}

interface FormInstance<T> {
  data: T
  get: () => T;
}


const nameData = { name: "levi" };
const formName: FormInstance<{ name: string }> = {
  data: nameData,
  get: () => nameData
};

const ageData = { age: 18 };
const formAge: FormInstance<{ age: number }> = {
  data: ageData,
  get: () => ageData
};


const dataName = tableSubmit(formName, data => ({ ...data, site: 'sh' }));
const dataAge = tableSubmit(formAge, undefined);

// remove undenfind args is any
const dataError = tableSubmit(formAge);

方法 2：类型推导 https://tsplay.dev/WPQG5w

function tableSubmit <T, R extends (((data: T) => any) | undefined)>(form: FormInstance<T>, filter?: R) {
  const params = form.get();
  const data  = filter !== undefined ? filter(params) : params;
  return data as undefined extends R ? T : R extends (...args: any) => any ? ReturnType<R> : T 
}

interface FormInstance<T> {
  data: T
  get: () => T;
}


const nameData = { name: "levi" };
const formName: FormInstance<{ name: string }> = {
  data: nameData,
  get: () => nameData
};

const ageData = { age: 18 };
const formAge: FormInstance<{ age: number }> = {
  data: ageData,
  get: () => ageData
};


const dataName = tableSubmit(formName, data => ({ ...data, site: 'sh' }));
const dataAge = tableSubmit(formAge, undefined);

const dataError = tableSubmit(formAge);

推荐用重载，非重载会把 void 改造成 undefined