对象是否在 javascript 深拷贝或浅拷贝中推入数组？
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Do objects pushed into an array in javascript deep or shallow copy?
提问by Travis J
Pretty self evident question...When using .push() on an array in javascript, is the object pushed into the array a pointer (shallow) or the actual object (deep) regardlessof type.
非常不言自明的问题...当在 javascript 中的数组上使用 .push() 时，无论类型如何，将对象推入数组是指针（浅）还是实际对象（深）。
回答by jfriend00
It depends upon what you're pushing. Objects and arrays are pushed as a pointer to the original object . Built-in primitive types like numbers or booleans are pushed as a copy. So, since objects are not copied in any way, there's no deep or shallow copy for them.
这取决于你在推动什么。对象和数组作为指向原始对象的指针推送。数字或布尔值等内置原始类型作为副本推送。因此，由于对象不会以任何方式复制，因此它们没有深拷贝或浅拷贝。
Here's a working snippet that shows it:
这是一个显示它的工作片段：
var array = [];
var x = 4;
let y = {name: "test", type: "data", data: "2-27-2009"};
// primitive value pushes a copy of the value 4
array.push(x); // push value of 4
x = 5; // change x to 5
console.log(array[0]); // array still contains 4 because it's a copy
// object reference pushes a reference
array.push(y); // put object y reference into the array
y.name = "foo"; // change y.name property
console.log(array[1].name); // logs changed value "foo" because it's a reference
// object reference pushes a reference but object can still be referred to even though original variable is no longer within scope
if (true) {
let z = {name: "test", type: "data", data: "2-28-2019"};
array.push(z);
}
console.log(array[2].name); // log shows value "test" since the pointer reference via the array is still within scope
回答by ruffin
jfriend00 is right on the mark here, but one small clarification: That doesn't mean you can't change what your variable is pointing to. That is, yinitially references some variable that you put into the array, but you can then take the variable named y, disconnect it from the object that's in the array now, and connect y(ie, make it reference) something different entirely without changing the object that now is referenced only by the array.
jfriend00 在这里恰到好处，但有一个小小的澄清：这并不意味着您不能更改您的变量所指向的内容。也就是说，y最初引用了您放入数组中的某个变量，但是您随后可以获取名为 的变量y，将其与现在数组中的对象断开连接，并连接y（即，使其引用）完全不同的东西，而无需更改对象现在仅由数组引用。
http://jsfiddle.net/rufwork/5cNQr/6/
http://jsfiddle.net/rufwork/5cNQr/6/
var array = [];
var x = 4;
var y = {name: "test", type: "data", data: "2-27-2009"};
// 1.) pushes a copy
array.push(x);
x = 5;
document.write(array[0] + "<br>"); // alerts 4 because it's a copy
// 2.) pushes a reference
array.push(y);
y.name = "foo";
// 3.) Disconnects y and points it at a new object
y = {};
y.name = 'bar';
document.write(array[1].name + ' :: ' + y.name + "<br>");
// alerts "foo :: bar" because y was a reference, but then
// the reference was moved to a new object while the
// reference in the array stayed the same (referencing the
// original object)
// 4.) Uses y's original reference, stored in the array,
// to access the old object.
array[1].name = 'foobar';
document.write(array[1].name + "<br>");
// alerts "foobar" because you used the array to point to
// the object that was initially in y.
