本文主要内容是vue3中的patchKeyedChildren函数，该函数为vue3diff最核心的方法，它的作用是将有key的children列表进行diff比较，最终以老children 为模版，通过patch move unmounted 完成diff
// can be all-keyed or mixed
const patchKeyedChildren = (
c1: VNode[],
c2: VNodeArrayChildren,
container: RendererElement,
parentAnchor: RendererNode | null,
parentComponent: ComponentInternalInstance | null,
parentSuspense: SuspenseBoundary | null,
isSVG: boolean,
slotScopeIds: string[] | null,
optimized: boolean
) => {
let i = 0
const l2 = c2.length
let e1 = c1.length - 1 // prev ending index
let e2 = l2 - 1 // next ending index
// 1. sync from start
// 从头开始比较
// (a b) c
// (a b) d e
while (i <= e1 && i <= e2) {
const n1 = c1[i]
const n2 = (c2[i] = optimized
? cloneIfMounted(c2[i] as VNode)
: normalizeVNode(c2[i]))
// 比较c1[i] c2[i] 是否值得比较，也就是是不是同一个元素 如果不是直接跳出循环
if (isSameVNodeType(n1, n2)) {
patch(
n1,
n2,
container,
null,
parentComponent,
parentSuspense,
isSVG,
slotScopeIds,
optimized
)
} else {
break
}
i++
}
// 2. sync from end
// 从尾开始比较
// a (b c)
// d e (b c)
while (i <= e1 && i <= e2) {
const n1 = c1[e1]
const n2 = (c2[e2] = optimized
? cloneIfMounted(c2[e2] as VNode)
: normalizeVNode(c2[e2]))
// 比较c1[i] c2[i] 是否值得比较，也就是是不是同一个元素 如果不是直接跳出循环
if (isSameVNodeType(n1, n2)) {
patch(
n1,
n2,
container,
null,
parentComponent,
parentSuspense,
isSVG,
slotScopeIds,
optimized
)
} else {
break
}
e1--
e2--
}
// 3. common sequence + mount
// (a b)
// (a b) c
// i = 2, e1 = 1, e2 = 2
// (a b)
// c (a b)
// i = 0, e1 = -1, e2 = 0
// 如果双端比较完后，新节点还有多，老节点没了即i > e1，那么，C2多出来的节点直接新增操作也就是执行patch：
if (i > e1) {
if (i <= e2) {
const nextPos = e2 + 1
const anchor = nextPos < l2 ? (c2[nextPos] as VNode).el : parentAnchor
while (i <= e2) {
patch(
null,
(c2[i] = optimized
? cloneIfMounted(c2[i] as VNode)
: normalizeVNode(c2[i])),
container,
anchor,
parentComponent,
parentSuspense,
isSVG,
slotScopeIds,
optimized
)
i++
}
}
}
// 4. common sequence + unmount
// (a b) c
// (a b)
// i = 2, e1 = 2, e2 = 1
// a (b c)
// (b c)
// i = 0, e1 = 0, e2 = -1
// 如果双端比较完后，新节点没有了，老节点有多余的，那么，C1多出来的节点直接删除即可也就是执行unmount：
else if (i > e2) {
while (i <= e1) {
unmount(c1[i], parentComponent, parentSuspense, true)
i++
}
}
// 5. unknown sequence
// [i ... e1 + 1]: a b [c d e] f g
// [i ... e2 + 1]: a b [e d c h] f g
// i = 2, e1 = 4, e2 = 5
// 两端的节点比较完，剩下中间的节点是乱序的即剩余的节点的两头两尾都不相同是乱序的
else {
const s1 = i // prev starting index
const s2 = i // next starting index
// 5.1 build key:index map for newChildren
// 构建一个以新节点的key为map的key,在数组中的索引位置为值的map，使用空间换时间的做法，在遍历一个数组的时候，直接在map中查询，降低时间复杂度。
const keyToNewIndexMap: Map<string | number | symbol, number> = new Map()
for (i = s2; i <= e2; i++) {
const nextChild = (c2[i] = optimized
? cloneIfMounted(c2[i] as VNode)
: normalizeVNode(c2[i]))
if (nextChild.key != null) {
if (__DEV__ && keyToNewIndexMap.has(nextChild.key)) {
warn(
`Duplicate keys found during update:`,
JSON.stringify(nextChild.key),
`Make sure keys are unique.`
)
}
keyToNewIndexMap.set(nextChild.key, i)
}
}
// 5.2 loop through old children left to be patched and try to patch
// matching nodes & remove nodes that are no longer present
let j
let patched = 0
const toBePatched = e2 - s2 + 1
let moved = false
// used to track whether any node has moved
let maxNewIndexSoFar = 0
// works as Map<newIndex, oldIndex>
// Note that oldIndex is offset by +1
// and oldIndex = 0 is a special value indicating the new node has
// no corresponding old node.
// used for determining longest stable subsequence
// 创建一个新数组乱序部分为长度的数组
const newIndexToOldIndexMap = new Array(toBePatched)
for (i = 0; i < toBePatched; i++) newIndexToOldIndexMap[i] = 0
// 遍历老数组在新数组中查找是否有key值相同的节点
for (i = s1; i <= e1; i++) {
const prevChild = c1[i]
// patched为值得比较的次数,如果大于toBePatched(新数组乱序部分的长度)说明新节点已经比较完毕，那么剩下的老节点都需要删除
if (patched >= toBePatched) {
// all new children have been patched so this can only be a removal
unmount(prevChild, parentComponent, parentSuspense, true)
continue
}
let newIndex
// 如果新节点有key值，那么就在map中查找是否有相同的key值，newIndex为在新数组中的索引
if (prevChild.key != null) {
newIndex = keyToNewIndexMap.get(prevChild.key)
} else {
// key-less node, try to locate a key-less node of the same type
// 如果新节点没有key值，那么就遍历新数组，寻找和老节点类型相同的节点
for (j = s2; j <= e2; j++) {
if (
newIndexToOldIndexMap[j - s2] === 0 // newIndexToOldIndexMap[j - s2] === 0，说明没有进行过patch
&&
isSameVNodeType(prevChild, c2[j] as VNode)
) {
newIndex = j
break
}
}
}
// 在新数组中没有老节点类型相同的节点，那么就直接删除老节点
if (newIndex === undefined) {
unmount(prevChild, parentComponent, parentSuspense, true)
} else {
// 保存下所有（值得比较也就是是相同的节点）的老节点的索引
newIndexToOldIndexMap[newIndex - s2] = i + 1
// 得到的索引newIndex是新节点在新数组中的索引，应该是一直增大的，如果newIndex < maxNewIndexSoFar说明需要移动
if (newIndex >= maxNewIndexSoFar) {
maxNewIndexSoFar = newIndex
} else {
moved = true
}
patch(
prevChild,
c2[newIndex] as VNode,
container,
null,
parentComponent,
parentSuspense,
isSVG,
slotScopeIds,
optimized
)
patched++
}
}
// 5.3 move and mount
// generate longest stable subsequence only when nodes have moved
// 如果新节点有移动，那么就需要生成最长的递增子序列 [1,8,6,3,4,5]=>(最长的递增子序列)[1,3,4,5]=>(最长的递增子序列在原数组中的索引)[0,3,4,5]
const increasingNewIndexSequence = moved
? getSequence(newIndexToOldIndexMap)
: EMPTY_ARR
j = increasingNewIndexSequence.length - 1
// looping backwards so that we can use last patched node as anchor
// 遍历新数组乱序部分，以老数组为模版，将新节点通过移动变为新数组中的顺序
for (i = toBePatched - 1; i >= 0; i--) {
const nextIndex = s2 + i
const nextChild = c2[nextIndex] as VNode
const anchor =
nextIndex + 1 < l2 ? (c2[nextIndex + 1] as VNode).el : parentAnchor
if (newIndexToOldIndexMap[i] === 0) {
// mount new
patch(
null,
nextChild,
container,
anchor,
parentComponent,
parentSuspense,
isSVG,
slotScopeIds,
optimized
)
} else if (moved) {
// move if:
// There is no stable subsequence (e.g. a reverse)
// OR current node is not among the stable sequence
if (j < 0 || //j<0有两种情况：1.最长递增子序列的长度为0，2.最长递增子序列的元素已经遍历完毕
i !== increasingNewIndexSequence[j] // 判断不为同一个元素
) {
move(nextChild, container, anchor, MoveType.REORDER)
} else {
// 不需要移动
j--
}
}
}
}
}
网友评论