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);Run Code Online (Sandbox Code Playgroud)
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?