Javascript 关闭异常

Question

The output from this script is: [ [ 1, 2, 3, 4 ], 10 ] [ [ 91, 92, 93, 94 ], 10 ] [ [ 8888, 8888, 8888, 8888 ], 10 ] [ [ 'one', 'two', 'three', 'four' ], 10 ]

But if I un-comment hogs = [2, 4, 6, 8], the output is: [ [ 1, 2, 3, 4 ], 10 ] [ [ 1, 2, 3, 4 ], 10 ] [ [ 1, 2, 3, 4 ], 10 ] [ [ 1, 2, 3, 4 ], 10 ]

If I leave it commented out but un-comment hogs = [333, 444, 555, 666], the output is: [ [ 1, 2, 3, 4 ], 10 ] [ [ 91, 92, 93, 94 ], 10 ] [ [ 8888, 8888, 8888, 8888 ], 10 ] [ [ 8888, 8888, 8888, 8888 ], 10 ]

I would like to understand simple javascript closures well enough to predict their behavior. Is that possible short of analysing the specs for the engine? I would also like to know why re-defining an entire array in one statement has such drastically different side effects from re-assigning the array's individual values one at a time. I am afraid to get very creative because I don't know how to predict side effects, and I can't find a clue in the literature.

var x = 10;
var hogs = [1, 2, 3, 4];
var array5 = (function () {
    var ar = [];
    var z = x;
    var w = hogs;
    function g() {
        ar.push(w);
        ar.push(z);
        return ar;
    } return g;
})()();

console.log(array5);

//hogs = [2, 4, 6, 8];  Uncommenting this prevents array5 from changing.     
x = 40;
hogs[0] = 91;
hogs[1] = 92;
hogs[2] = 93;
hogs[3] = 94;
console.log(array5);

hogs[0] = 8888;
hogs[1] = 8888;The output from this script is:
hogs[2] = 8888;
hogs[3] = 8888;
console.log(array5);

// hogs = [333, 444, 555, 666]; Un-commenting this prevents.. 

hogs[0] = 'one';
hogs[1] = 'two';
hogs[2] = 'three';
hogs[3] = 'four';
x = 40;
console.log(array5);

I don't think you were being repetitive; you elegantly put the whole thing in a nutshell when you wrote "You are not saying 'tell the variable w to reference whatever object hogs references". What you are saying is "tell the variable w to reference whatever object hogs references at this current moment'" You were kind to take the time to explain this in detail. Perhaps you empathized with someone learning the language who simplistically imagined that the search for a variable value proceeded outwards through the scope chain all the way to Object object if necessary, while in this case the value in the enclosing function is ignored until the value in global scope is obliterated by a new definition. It is good to know that changing the values in an array maintains the array itself, intact but with some new or changed value assignments; and that is very very different from redefining an array, making the variable point to something entirely new. Images of some objects in memory standing still while others disappear and reappear elsewhere in memory dance in my head; which rhymes with "time for bed." For anyone else who might be interested in this, the analogous situation seems to exist for all objects, as illustrated by the following code:

enter code here
var x = 10;
var hogs = {'a': 1, 'b': 2, 'c' : 3, 'd' : 4};
var obj = {};
obj.h = hogs;

var ob5 = (function (arg) {
    function g() {
        var w = arg;
        return w;
    } return g;
})(hogs)();

console.log(ob5);

//hogs = [2, 4, 6, 8];  Uncommenting this prevents array5 from  changing.
x = 40;
hogs['a'] = 91;
hogs['b'] = 92;
hogs['c'] = 93;
hogs['d'] = 94;
console.log(ob5);

hogs.a = 8888;
hogs.b = 8888;
hogs.c = 8888;
hogs.d = 8888;
console.log(ob5);

hogs = {'a':333, 'b':444, 'c':555, 'd':666}; // Prevents further changes.

hogs.a = 'one';
hogs.b = 'two';
hogs.c = 'three';
hogs.d = 'four';
x = 40;
console.log(ob5);

Answer 1

闭包在应用(调用)时返回 hogs 的当前值。如果稍后更新 hogs 以指向新数组，array5(调用闭包的结果)仍引用旧副本。您的更新仅反射(reflect)在 hogs 的当前副本中，因此无论您更改什么地方，您的更新都将反射(reflect)在 array5 中.

您可以再次应用闭包，您将获得一个反射(reflect) hogs 当前值的新值，但您必须为其指定一个名称(将其放入变量中)以供引用又是它。

---

That's what I thought, and what I would have predicted. But I don't get that predicted behavior unless I include (un-comment) the line: hogs = [2, 4, 6, 8]. That freezes array5 at its value prior to "hogs = [2, 4, 6, 8]." Otherwise, array5 acts like a reference to the current value of hogs. I don't know how that behavior might have been predicted, and I can't explain it after the fact. Does it make sense to you?

是的，这就是您应该期望它的行为方式。数组是可变对象，因此在一个地方更改数组将反射(reflect)在您更改它的任何其他地方。您所做的实际上与此没有太大区别:

var x = [1, 2, 3];
var y = x;
var x[0] = 4;
// What would you expect the value of y to be here?

如果您真的想要一个数组的副本，那么您应该使用slice() 方法进行显式复制:

var array5 = (function () {
    var ar = [];
    var z = x;
    var w = hogs.slice(); // <== Explicit copy!
    function g() {
        ar.push(w);
        ar.push(z);
        return ar;
    } return g;
})()();