问题
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) regardless of type.
回答1:
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.
Here's a working snippet that shows it:
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2:
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, y
initially 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.
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.
来源:https://stackoverflow.com/questions/8660901/do-objects-pushed-into-an-array-in-javascript-deep-or-shallow-copy