JavaScript closure的一个问题

function f1() {
    var n = 999;
    nAdd = function() {
        n += 1
    }
    function f2() {
        alert(n);
    }
    return f2;
}
var result1 = f1();
var result2 = f1();
var result3 = f1();
nAdd();
result1();  //999
result2(); //999
result3(); //1000

以上是困惑的问题？为什么nAdd（）影响最近的闭包。
stackoverflow的两个答案，合起来能很好的理解这个问题。

1.进一步明白细节的回答

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It has to do with when nAdd is assigned the function. Note that when you create a closure, a new “copy” (for lack of a better word) of the local variables are created. Thus, result1’s n is different from result2s n which is different from result3’s n. They are seperate, and each closure cannot access another closure’s n.

Look at this line:

nAdd = function() {
     n += 1;
}

This assigns nAdd a new closure each time. Each time, this closure will only affect the most recent “copy” of n.
So when you are doing.

var result1 = f1(); // assign nAdd for the first time, referring to result1's n.
var result2 = f1(); // re-assign nAdd, now it affects result2's n.
var result3 = f1(); // re-assign nAdd, now it affect result3's n.

nAdd got assigned a new closure each time. The last time, nAdd got assigned a closure with result3’s copy of n.
Thus, when you do nAdd(), you only increment result3’s n.
Here’s an example that might clear things up.

function f1() {
    var n = 999;
    nAdd = function() {
        n += 1
    }
    function f2() {
        alert(n);
    }
    return f2;
}
var result1 = f1();
var nAdd1 = nAdd;
var result2 = f1();
var nAdd2 = nAdd;
var result3 = f1();
var nAdd3 = nAdd;
nAdd3();
result1();  //999
result2(); //999
result3(); //1000
nAdd1();
result1(); // 1000
nAdd1();
result1(); // 1001
nAdd2();
result2(); // 1000
nAdd();
result3(); // 1001 (the most recent nAdd result3's n).

To further elaborate, consider what would happen if you did this instead:

var result1 = f1();
nAdd();
var result2 = f1();
var result3 = f1();
result1(); // 1000
result2(); // 999
result3(); //999

Or this:

var result1 = f1();
var result2 = f1();
nAdd();
var result3 = f1();
result1(); // 999
result2(); // 1000
result3(); // 999

It becomes obvious that nAdd updates only the most recent invocation’s n!

2.调用本质的回答

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First, you need to be clear about the following thing:

nAdd is a global scoped function. But it is using a local scoped variable n.

Every time you re-create f1(), the pointer of nAdd function will change.

Finally it will changed to the closest f1() closure and the n variable inside nAdd function will also point to the closest f1() closure. So it can only change the value of the closest one.

在这里插入代码片