我以为我有一个this引用传递到Javascript闭包中的私有函数，我想错了

我认为代码解释得很好。我使用闭包来保持称为reverseAll私有的方法。reverseAll是递归的，但我认为你不需要担心这个。我试图在reverseAll函数中引用this.head，但我发现它是undefined。所以我传递了一个对this.head的引用，并通过递归调用继续传递它。唉，这不是命中注定的。我可以把reverseAll方法从闭包中拉出来，我将有一个对this.head的引用。但我想知道为什么我传递的this.head引用不是"引用的副本"或"指针的副本"，如果你喜欢Javascript，当你将对象传递给函数时。顺便说一下，this.head是node对象。

这里是代码(引用jQuery是因为Stackoverflow片段"IDE"失败时，行var obj = new LinkedList();，所以我添加了一个文档。准备好了，jQuery不需要任何其他原因:

function LinkedList() {
    this.head = null;
};
LinkedList.prototype = (function () {
    function reverseAll(current, prev, theHead) {
        if (!current.next) { //we have the head
            console.log('ending recursion, new head!!');
            console.log('we have a refence to this.head in theHead or so I thought:');
            console.log(theHead);
            theHead = current;
            theHead.next = prev;
            console.log('theHead has a new "pointer":');
            console.log(theHead);
            return;
        }
        var next = current.next;
        current.next = prev;
        //keep passing the theHead reference through recursion
        reverseAll(next, current, theHead);
    };
    return {
        constructor: LinkedList,
        reverse: function () {
            console.log('clone head to iterate and change');
            console.log('but also pass in reference of this.head obj as this.head is a node obj and this.head will be undefined in reverseAll()');
            var headClone = JSON.parse(JSON.stringify(this.head));
            reverseAll(headClone, null, this.head);
        }
    }
})();
LinkedList.prototype.add = function(value) {
    var node = {
        value: value,
        next: null
    };
    var current;
    if (this.head === null) {
        this.head = node;
    } else {
        current = this.head;
        while (current.next) {
            current = current.next;
        }
        current.next = node;
    }
    return node;
}
LinkedList.prototype.remove = function(node) {
    var current, value = node.value;
    if (this.head !== null) {
        if (this.head === node) {
            this.head = this.head.next;
            node.next = null;
            return value;
        }
        //find node if node not head
        current = this.head;
        while (current.next) {
            if (current.next === node) {
                current.next = node.next;
                return value;
            }
            current = current.next;
        }
    }
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
$(function() {
    var obj = new LinkedList();
    
    for (var i = 1; i <= 10; i++) {
        obj.add(i);
    }
    console.log('about to call obj.reverse()!');
    console.log('obj.head:');
    console.log(obj.head);
    obj.reverse();
    console.log('obj instance obj.head after reverse call, it has not changed!!:');
    console.log(obj.head);
});
</script>