我注意到调用 .map() 而不将其分配给变量会使其返回整个数组，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工人，为什么雇员也会发生变异？
发生这种情况是因为数组中的对象仍然由相同的指针引用。 （您的数组仍然引用内存中的相同对象）。还，Array.prototype.map()总是返回一个数组，它的结果应该分配给一个变量，因为它不这样做in-place映射。当您更改对象的属性时map方法，你应该考虑使用.forEach()相反，修改复制的员工数组中对象的属性。要制作员工数组的副本，您可以使用以下命令：
const workers = JSON.parse(JSON.stringify(employees));
请参阅下面的示例：
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，您需要使用另一种方法来深复制
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