在这段代码中，我obj使用展开运算符复制到copiedObj. 然后将checked属性值修改为false. 但是，当 I 时console.log(obj.checked[0])，它返回 false 而不是 true。而且似乎中的checked值copiedObj也改变了原始属性值。这是为什么？
const obj = [{id: 1, checked: true}];
const copiedObj = [...obj];
copiedObj.checked[0] = false;
console.log(obj.checked[0]);obj和copiedObj是数组，而不是普通对象。更改状态copiedObj（通过checked向其添加属性）不会更改状态，obj因为它们是单独的数组。
但是，两个数组都包含对同一个对象的引用（上面有一个cheecked）。所以如果你这样做：
checkedObj[0].checked = true;
这会改变那个对象的状态，你会看到你是否在obj[0]或上查找该对象checkedObj[0]。
如果要进行深拷贝以便拥有单独的数组和单独的对象，请参阅此问题的答案。
由于我 99% 确定你的意思checkedObj[0].checked = true，我将解释这段代码中发生的事情：
// Creates an array containing an object
const obj = [{id: 1, checked: true}];
// Creates a new array that contains the *same* object (NOT a *copy* of it)
const copiedObj = [...obj];
// Sets `checked` on that one object that is in both arrays
copiedObj[0].checked = false;
// Looks at the `checked` property on that one object that is in both arrays
console.log(obj[0].checked);一步步：
后
// Creates an array containing an object
const obj = [{id: 1, checked: true}];
在记忆中你有类似的东西
+−−−−−−−−−−−−−−−+
obj:Ref44329−−−−−>| (数组) |
+−−−−−−−−−−−−−−+ +−−−−−−−−−−−−−−−+
| 0：Ref82445 |−−−>| (对象) |
+−−−−−−−−−−−−−−+ +−−−−−−−−−−−−−−−+
| 编号：1 |
| 检查： 真 |
+−−−−−−−−−−−−−−−−−+
然后当你做
// Creates a new array that contains the *same* object (NOT a *copy* of it)
const copiedObj = [...obj];
你有这样的事情：
+−−−−−−−−−−−−−−−+
obj:Ref44329−−−−−−−−−−−>| (数组) |
+−−−−−−−−−−−−−−−+
| 0：Ref82445 |−−+
+−−−−−−−−−−−−−−+ |
|
| +−−−−−−−−−−−−−−−−−−+
+−−>| (对象) |
+−−−−−−−−−−−−−−+ | +−−−−−−−−−−−−−−−−−−+
checkedObj:Ref12987−−−>| (数组) | | | 编号：1 |
+−−−−−−−−−−−−−−+ | | 检查： 真 |
| 0: Ref82445 |−−+ +−−−−−−−−−−−−−−−−+
+−−−−−−−−−−−−−−+
然后当你做
// Sets `checked` on that one object that is in both arrays
copiedObj[0].checked = false;
它改变了两个数组指向的对象
+−−−−−−−−−−−−−−−+
obj:Ref44329−−−−−−−−−−−>| (数组) |
+−−−−−−−−−−−−−−−+
| 0：Ref82445 |−−+
+−−−−−−−−−−−−−−+ |
|
| +−−−−−−−−−−−−−−−−−−+
+−−>| (对象) |
+−−−−−−−−−−−−−−+ | +−−−−−−−−−−−−−−−−−−+
已检查对象：Ref12987−−−>| (数组) | | | 编号：1 |
+−−−−−−−−−−−−−−+ | | 检查：假|
| 0：Ref82445 |−−+ +−−−−−−−−−−−−−−−−+
+−−−−−−−−−−−−−−−+
...因此false无论您通过哪个数组查看它，查找它都会给您。
obj 是一个对象数组，当你这样做时：
const copiedObj = [...obj];
您创建了一个新数组，但在该数组中您仍将引用来自 obj
发生这种情况是因为您有两个级别：数组 -> 对象，但扩展运算符仅适用于第一级（数组）
这是因为当您copiedObj通过扩展运算符创建 时，您仍在引用最初在内存中创建的对象。
为了解决这个问题，你会得到预期的行为，也是创建对象内所有属性的另一个浅拷贝。
ps 我修改copiedObj[0].checked = false; console.log(obj[0].checked)是因为我认为您试图选择您创建的数组中的第一个元素。
const obj = [{id: 1, checked: true}];
const copiedObj = [Object.assign({}, obj[0])];
copiedObj[0].checked = false;
console.log(obj[0].checked)
//returns true