Async/Await中如何优雅地退出不确定时间回调函数？

const a = (): promise<void> => {
  // 监听按钮的属性变化
  return new promise((resolve) => {
    const callback = (mutations: mutationrecord[]) => {
      // 退出
      resolve();
    };
    const observer = new mutationobserver(callback);
    observer.observe(buttonel, {
      attributes: true,
    });

    // 触发按钮点击事件并监听属性变化
    p().then(() => {
      buttonel.click();
    });
  });
};

const b = async (): promise<void> => {
  // 监听按钮的属性变化，如何退出？
  const callback = (mutations: mutationrecord[]) => {};
  const observer = new mutationobserver(callback);
  observer.observe(buttonel, {
    attributes: true,
  });

  // 触发按钮点击事件并监听属性变化
  await p();
  buttonel.click();
};

const b = async (): Promise<void> => {
  // 创建 Promise 实例
  const { promise, resolve } = Promise.withResolvers<void>();

  // 设置回调和观察器
  const callback = (mutations: MutationRecord[]) => {
    resolve();
  };
  const observer = new MutationObserver(callback);
  observer.observe(buttonEl, {
    attributes: true,
  });

  // 触发按钮点击事件并监听属性变化
  await p();
  buttonEl.click();

  // 使用 await promise 或 return promise 返回 Promise
  await promise;
};