fix: useMemoizedFn return a plain function type

[fanck0605]

[中文版模板 / Chinese template]

🤔 This is a ...

• Bug fix

💡 Background and solution

我们考察如下代码

export default function App() {
  const fn = () => {
    console.log("hello");
  };

  fn.hello = "hello"; // fn: {():void,hello: string}

  const memoizedFn = useMemoizedFn(fn); 

  return (
    <div>
      <div>{`${fn.hello}` /* hello */}</div>
      <div>{`${memoizedFn.hello}` /* undefined */}</div>
    </div>
  );
}

TypeScript 推导出 memoizedFn.hello 为不可空的 string, 但实际访问则是 undefined。
如果 memoizedFn.hello 还是个对象，继续使用 . 运算符，那么还会有更可怕的异常发生。

hooks/packages/hooks/src/useMemoizedFn/index.ts

Lines 20 to 22 in cbc8f8f

	memoizedFn.current = function (this, ...args) {
	return fnRef.current.apply(this, args);
	} as T;

原来的代码，由于使用的类型强制转换，导致类型推导出现错误，我们这里的返回值类型，应该剔除 fn 对象中的所有属性，而只保留其函数的部分。

因此我们可以定义一个 ts type，它的泛型参数可以接受任意函数对象，然后转换成一个普通的函数

type PickFunction<T extends noop> = (
  this: ThisParameterType<T>,
  ...args: Parameters<T>
) => ReturnType<T>;

📝 Changelog

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🇨🇳 Chinese

☑️ Self Check before Merge

⚠️ Please check all items below before review. ⚠️

• Doc is updated/provided or not needed
• Demo is updated/provided or not needed
• TypeScript definition is updated/provided or not needed
• Changelog is provided or not needed

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❌ fanck0605
❌ brickspert
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[awmleer]

LGTM