In the last part, we gave our javascript structure creator function to ability to give each node a hashcode.

We did this so the comparison function can compare a node’s children, to rearrage them rather than obliterate them.

The first thing we’ll do is get the hashcodes before we look at the node’s children. v and v1 equate to the two virtual dom trees we’re comparing.

var v_hashes = v.children.map(c => c.hashcode )
var v1_hashes = v1.children.map(c => c.hashcode )

Next we’ll make an array of the positions of the v1 hashes and whether they exist in the v hashes using indexOf().

Given the v hash array [1111, 2222] and the v1 hash array [3333, 1111, 2222] we’ll get [-1, 0, 1].

indexOf() is telling us: the zeroth item in v1 doesn’t exist in v but the first is the zeroth element, and the second is the first element.

We’re also grabbing v1’s hash in the returning array for later reference.

var positions = v1_hashes.map((h, i) => v_hashes.indexOf(h) )

With this new structure we will loop over it. It will tell us if there’s a new item in the v1 virtual dom that’s not in the v virtual dom, i.e. we find a -1 value.

positions.forEach((index_in_v, v1_pos) => {
  if(index_in_v != -1) return
  console.log("insert " + v1_hashes[v1_pos] + " into " + v1_pos + " at " + path)
  v.children.splice(v1_pos, 0, JSON.parse(JSON.stringify(v1.children[v1_pos])))
})

So we look for that -1, and then rearrange the original v array. This means, later, when we compare the two tree branches our compare function will see they’re the same.

Of course, our compare function must output the fact – here indicated initially as the console.log – that the DOM manipulation function must actually do this rearragement of the DOM.

Although there is much more to do with this function, we can now fix our previous problem with the algorithm:

The v virtual DOM tree was:

[{
    "type": "node",
    "name": "DIV",
    "attrs": [],
    "children": [{
        "type": "node",
        "name": "P",
        "attrs": [],
        "children": [{
            "type": "text",
            "value": "original",
            "children": [],
            "hashcode": -68148979
        }],
        "hashcode": -1072170396
    }],
    "hashcode": -154850033
}]

And v1 was:

[{
    "type": "node",
    "name": "DIV",
    "attrs": [],
    "children": [{
        "type": "node",
        "name": "P",
        "attrs": [],
        "children": [{
            "type": "text",
            "value": "new",
            "children": [],
            "hashcode": 1156737192
        }],
        "hashcode": -1348153726
    }, {
        "type": "node",
        "name": "P",
        "attrs": [],
        "children": [{
            "type": "text",
            "value": "original",
            "children": [],
            "hashcode": -68148979
        }],
        "hashcode": -1072170396
    }],
    "hashcode": 1511584547
}]

In other words, v1 had a new P node (hashcode: -1348153726) above the old one.

And this function:

compare = function(v, v1, path) {
    var v_hashes = v.children.map(c => c.hashcode )
    var v1_hashes = v1.children.map(c => c.hashcode )
    var positions = v1_hashes.map((h, i) => v_hashes.indexOf(h) )
    positions.forEach((index_in_v, v1_pos) => {
      if(index_in_v != -1) return
      console.log("insert " + v1_hashes[v1_pos] + " into " + v1_pos + " at " + path)
      v.children.splice(v1_pos, 0, JSON.parse(JSON.stringify(v1.children[v1_pos])))
    })
    v.children.forEach((_, i) => compare(v.children[i], v1.children[i], path + "" + i) )
}; 
compare(JSON.parse(JSON.stringify(vd[0])), JSON.parse(JSON.stringify(vd1[0])), "")

Will output only.

insert -1348153726 into 0

Next up, we’ll look at how we can remove nodes.