javascript中异步延迟初始化的规范安全方法是什么？

Question

我的程序中有这种惰性初始化代码：

let user = null;        
let getUser = async () => {
  if(!user) {
    user = await getUserSomehow();
  }
  return user;
};

我知道这是不安全的，因为如果我有下一个代码可能会出现竞争情况：

// one place of the program
let u1 = await getUser();
...
// another place of the program running during getUserSomehow() for u1 still hasn't finished
let u2 = await getUser();

getUserSomehow() 将被调用两次而不是一次。

如何避免这种情况？

Answer 1

第一次调用时，分配一个 Promise，而在进一步调用时，返回该 Promise：

let userProm = null;
let getUser = () => {
  if (!userProm) {
    userProm = getUserSomehow();
  }
  return userProm;
};

更好的是，将userProm 的范围仅限于getUser，这样更安全、更清晰：

const getUser = (() => {
  let userProm = null;
  return () => {
    if (!userProm) {
      userProm = getUserSomehow();
    }
    return userProm;
  };
})();

const getUserSomehow = () => {
  console.log('getting user');
  return Promise.resolve('data');
};

const getUser = (() => {
  let userProm = null;
  return () => {
    if (!userProm) {
      userProm = getUserSomehow();
    }
    return userProm;
  };
})();

(async () => {
  const userProm1 = getUser();
  const userProm2 = getUser();
  Promise.all([userProm1, userProm2]).then(() => {
    console.log('All done');
  });
})();

您现有的代码碰巧是安全的，因为对user 的分配将在getUser 的第一次调用完成之前发生：

const getUserSomehow = () => {
  console.log('Getting user');
  return Promise.resolve('data');
};

let user = null;
let getUser = async() => {
  if (!user) {
    user = await getUserSomehow();
  }
  return user;
};

(async () => {
  let u1 = await getUser();
  let u2 = await getUser();
  console.log('Done');
})();

但如果 Promise 是在其中一个被 awaited 完成之前并行初始化的，则不会：

const getUserSomehow = () => {
  console.log('Getting user');
  return Promise.resolve('data');
};

let user = null;
let getUser = async() => {
  if (!user) {
    user = await getUserSomehow();
  }
  return user;
};

(async() => {
  let u1Prom = getUser();
  let u2Prom = getUser();
  await Promise.all([u1Prom, u2Prom]);
  console.log('Done');
})();

如上所示，将 Promise 分配给持久变量（而不是 awaiting getUser 中的值）可以解决此问题。