javascript - 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，改为分配一个 Promise，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，范围 userProm仅在 getUser 内，更安全和清晰:

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');
  });
})();

您现有的代码恰好是安全的，因为分配给 user将在第一次调用 getUser 之前发生完成:

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');
})();

但是，如果 Promise 是并行初始化的，则不会是在其中一个为 await 之前。首先完成:

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 await 中的值 getUser )可以解决此问题。