如何防止突变

Question

我一直在尝试编写一个计算混合账单的函数，而没有使用ES6的突变。
我在下面有一个代码，它的作用是，它需要一组人的钱，添加了所有代码，以获取总金额，然后循环，然后分配给用户的长度，以获取多少钱人。然后，我从用户数量中减去每人份额，因此列表包含负/正值或0，具体取决于它们贡献了多少总账单。然后，为了提取负数和阳性数量的人来说，使用大于0且小于0的人的用户过滤器。

到目前为止，代码正常运行，并且按照我的预期运行良好，但是我使用的是不断变化的变量，从我的理解中导致突变，是否有其他方法可以防止此代码中的突变，

您可以查看代码在摘要上

const users = [
  { id: 1, username: "John One", amount: 50 },
  { id: 2, username: "John Two", amount: 75 },
  { id: 3, username: "John Three", amount: 100 },
  { id: 4, username: "John Four", amount: 125 },
  { id: 5, username: "John five", amount: 150 },
  { id: 6, username: "John six", amount: 50 },
  { id: 7, username: "John seven", amount: 150 },
  { id: 8, username: "John eight", amount: 250 },
  { id: 9, username: "John nine", amount: 150 },
  { id: 10, username: "John Ten", amount: 100 },
]


const total = users.reduce((acc, user) => acc + user.amount, 0);
const perPersonShare = total / users.length;
const sortedUsers = users.map((user) => { return { ...user, amount: user.amount - perPersonShare, hasToPay: [] } });
const negativeAmount = sortedUsers.filter(user => user.amount < 0);
const positiveAmount = sortedUsers.filter(user => user.amount > 0);


const res = positiveAmount.reduce((acc, user) => {

  let balance = user.amount;

  return acc.map((ower) => {

    if (ower.amount === 0 || balance === 0) return ower;

    if (balance <= Math.abs(ower.amount)) {
      const updated = {
        ...ower,
        hasToPay: [...ower.hasToPay, { to: user.username, amount: balance }],
        amount: ower.amount + balance,
      };

      balance = 0;
      return updated;
    }

    if (balance >= Math.abs(ower.amount)) {
      const updated = {
        ...ower,
        hasToPay: [...ower.hasToPay, { to: user.username, amount: Math.abs(ower.amount) }],
        amount: 0,
      };

      balance = balance + ower.amount;
      return updated;
    }

    return ower;

  });


}, negativeAmount);

console.log(JSON.stringify(res, 0, 2))
console.log(perPersonShare);
console.log(total);

Answer 1

我是新手的，所以我的答案可能有效，但是您可以使用object.freeze（）冻结对象，而是使用数组来定义用户，而是可以使用对象定义它们。 It is from my understanding

here is some links I found they might be useful to you

https://medium.com/@fknussel/arrays-Arrays-abigns-and-mutations-6b23348b54aahttps://medium.comnyum.com/@fknussel/arrays-brays-objects-mutations-mutations-mutations-mutations-mutations-mutations-mutations-mutations-mutations-mutations-mutations-mutions-and-mutations-and-mutations-and-mutiations-mutim-6b2348b54aaaa