复制数组上的 Javascript .map()

Question

我注意到在不将其分配给变量的情况下调用 .map() 会使它返回整个数组，而不仅仅是更改的属性:

const employees = [{
    name: "John Doe",
    age: 41,
    occupation: "NYPD",
    killCount: 32,
  },
  {
    name: "Sarah Smith",
    age: 26,
    occupation: "LAPD",
    killCount: 12,
  },
  {
    name: "Robert Downey Jr.",
    age: 48,
    occupation: "Iron Man",
    killCount: 653,
  },

]

const workers = employees.concat();

workers.map(employee =>
  employee.occupation == "Iron Man" ? employee.occupation = "Philantropist" : employee.occupation
);

console.log(employees);

但考虑到 .concat() 创建了原始数组的副本并将其分配给 worker，为什么 employee 也会发生变异？

Answer 1

发生这种情况是因为数组中的对象仍被相同的指针引用。 (您的数组仍然引用内存中的相同对象)。此外，Array.prototype.map() 总是返回一个数组，它的结果应该分配给一个变量，因为它不进行就地 映射。当您在 map 方法中更改对象的属性时，您应该考虑改用 .forEach() 来修改复制的 employees 数组中对象的属性。要复制您的员工数组，您可以使用以下命令:

const workers = JSON.parse(JSON.stringify(employees));

请看下面的例子:

const employees = [
  {
    name: "John Doe",
    age: 41,
    occupation: "NYPD",
    killCount: 32,
  },
  {
    name: "Sarah Smith",
    age: 26,
    occupation: "LAPD",
    killCount: 12,
  },
  {
    name: "Robert Downey Jr.",
    age: 48,
    occupation: "Iron Man",
    killCount: 653,
  },

]


const workers = JSON.parse(JSON.stringify(employees));
workers.forEach(emp => {
  if(emp.occupation == "Iron Man") emp.occupation = "Philantropist";
});

console.log("--Employees--")
console.log(employees);
console.log("\n--Workers--");
console.log(workers);

• 注意:如果您的数据中有任何方法，您需要使用其他方法进行深拷贝