React中diff算法的理解#

{
    type: "div",
    props: {
        className: "root"
        name: "root",
        children: [{
            type: "p",
            props: {
                children: [{
                    type: "text",
                    props: {
                        text: "1"
                    }
                    
                }]
            }    
        },{
            type: "div",
            props: {
                children: [{
                    type: "text",
                    props: {
                        text: "11"
                    }
                }]
            }
        }]
    }
}

// react-reconciler/src/ReactChildFiber.js line 1246
export function reconcileChildren(
  current: Fiber | null,
  workInProgress: Fiber,
  nextChildren: any,
  renderExpirationTime: ExpirationTime,
) {
  if (current === null) { // 首次渲染 创建子节点的`Fiber`实例
    workInProgress.child = mountChildFibers(
      workInProgress,
      null,
      nextChildren,
      renderExpirationTime,
    );
  } else { // 否则调用`reconcileChildFibers`去做`diff`
    workInProgress.child = reconcileChildFibers(
      workInProgress,
      current.child,
      nextChildren,
      renderExpirationTime,
    );
  }
}

// react-reconciler/src/ReactChildFiber.js line 1370
export const reconcileChildFibers = ChildReconciler(true);
export const mountChildFibers = ChildReconciler(false);

function ChildReconciler(shouldTrackSideEffects) { 
  // ...
  function deleteChild(){
      // ...
  }
  function useFiber(){
      // ...
  }
  function placeChild(){
      // ...
  }
  function placeSingleChild(){
      // ...
  }
  function updateTextNode(){
      // ...
  }
  function updateElement(){
      // ...
  }
  function updatePortal(){
      // ...
  }
  function updateFragment(){
      // ...
  }
  function createChild(){
      // ...
  }
  function updateSlot(){
      // ...
  }
  function updateFromMap(){
      // ...
  }
  function warnOnInvalidKey(){
      // ...
  }
  function reconcileChildrenArray(){
      // ...
  }
  function reconcileChildrenIterator(){
      // ...
  }
  function reconcileSingleTextNode(){
      // ...
  }
  function reconcileSingleElement(){
      // ...
  }
  function reconcileSinglePortal(){
      // ...
  }
  function reconcileChildFibers(){
      // ...
  }
  return reconcileChildFibers;
}

if (currentFirstChild !== null && currentFirstChild.tag === HostText) {
  // We already have an existing node so let's just update it and delete
  // the rest.
  deleteRemainingChildren(returnFiber, currentFirstChild.sibling); // 删除兄弟
  const existing = useFiber(currentFirstChild, textContent, expirationTime);
  existing.return = returnFiber;
  return existing; // 复用
}

// The existing first child is not a text node so we need to create one
// and delete the existing ones.
// 创建新的Fiber节点，将旧的节点和旧节点的兄弟都删除 
deleteRemainingChildren(returnFiber, currentFirstChild);
const created = createFiberFromText(
  textContent,
  returnFiber.mode,
  expirationTime,
);

// react-reconciler/src/ReactChildFiber.js line 1132
const key = element.key;
let child = currentFirstChild;
while (child !== null) {
  // TODO: If key === null and child.key === null, then this only applies to
  // the first item in the list.
  if (child.key === key) {
    if (
      child.tag === Fragment
        ? element.type === REACT_FRAGMENT_TYPE
        : child.elementType === element.type ||
          // Keep this check inline so it only runs on the false path:
          (__DEV__
            ? isCompatibleFamilyForHotReloading(child, element)
            : false)
    ) {
      deleteRemainingChildren(returnFiber, child.sibling); // 因为当前节点是只有一个节点，而老的如果是有兄弟节点是要删除的，是多余的
      const existing = useFiber(
        child,
        element.type === REACT_FRAGMENT_TYPE
          ? element.props.children
          : element.props,
        expirationTime,
      );
      existing.ref = coerceRef(returnFiber, child, element);
      existing.return = returnFiber;
      // ...
      return existing;
    } else {
      deleteRemainingChildren(returnFiber, child);
      break;
    }
  } else {
    deleteChild(returnFiber, child); // 从child开始delete
  }
  child = child.sibling; // 继续从子节点中找到一个可复用的节点
}

// react-reconciler/src/ReactChildFiber.js line 1178
if (element.type === REACT_FRAGMENT_TYPE) {
  const created = createFiberFromFragment(
    element.props.children,
    returnFiber.mode,
    expirationTime,
    element.key,
  );
  created.return = returnFiber;
  return created;
} else {
  const created = createFiberFromElement(
    element,
    returnFiber.mode,
    expirationTime,
  );
  created.ref = coerceRef(returnFiber, currentFirstChild, element);
  created.return = returnFiber;
  return created;
}

// react-reconciler/src/ReactChildFiber.js line 756
function reconcileChildrenArray(
    returnFiber: Fiber,
    currentFirstChild: Fiber | null,
    newChildren: Array<*>,
    expirationTime: ExpirationTime,
  ): Fiber | null {
    // 机翻注释
    // 这个算法不能通过两端搜索来优化，因为我们在光纤上没有反向指针。我想看看我们能用这个模型走多远。如果最终不值得权衡，我们可以稍后再添加。
    // 即使是双端优化，我们也希望在很少有变化的情况下进行优化，并强制进行比较，而不是去寻找地图。它想探索在前进模式下首先到达那条路径，并且只有当我们注意到我们需要很多向前看的时候才去地图。这不能处理反转以及两个结束的搜索，但这是不寻常的。此外，要使两端优化在Iterables上工作，我们需要复制整个集合。
    // 在第一次迭代中，我们只需在每次插入/移动时都碰到坏情况（将所有内容添加到映射中）。
    // 如果更改此代码，还需要更新reconcileChildrenIterator()，它使用相同的算法。
    let oldFiber = currentFirstChild;
    let lastPlacedIndex = 0;
    let newIdx = 0;
    let nextOldFiber = null;
     // 第一个for循环，按照index一一对比，当新老节点不一致时退出循环并且记录退出时的节点及oldFiber节点
    for (; oldFiber !== null && newIdx < newChildren.length; newIdx++) {
      if (oldFiber.index > newIdx) { // 位置不匹配
        nextOldFiber = oldFiber;  // 下一个即将对比的旧节点
        oldFiber = null; // 如果newFiber也为null(不能复用)就退出当前一一对比的for循环
      } else {
        nextOldFiber = oldFiber.sibling; //正常的情况下 为了下轮循环，拿到兄弟节点下面赋值给oldFiber
      }
      // //如果节点可以复用(key值匹配)，就更新并且返回新节点，否则返回为null，代表节点不可以复用
      const newFiber = updateSlot( // 判断是否可以复用节点
        returnFiber,
        oldFiber,
        newChildren[newIdx],
        expirationTime,
      );
      // 节点无法复用 跳出循环
      if (newFiber === null) { 
        // TODO: This breaks on empty slots like null children. That's
        // unfortunate because it triggers the slow path all the time. We need
        // a better way to communicate whether this was a miss or null,
        // boolean, undefined, etc.
        if (oldFiber === null) {
          oldFiber = nextOldFiber; // 记录不可以复用的节点并且退出对比
        }
        break; // 退出循环
      }
      if (shouldTrackSideEffects) {
        // 没有复用已经存在的节点，就删除掉已经存在的节点
        if (oldFiber && newFiber.alternate === null) {
          // We matched the slot, but we didn't reuse the existing fiber, so we
          // need to delete the existing child.
          deleteChild(returnFiber, oldFiber);
        }
      }
      // 本次遍历会给新增的节点打 插入的标记
      lastPlacedIndex = placeChild(newFiber, lastPlacedIndex, newIdx);
      if (previousNewFiber === null) {
        // TODO: Move out of the loop. This only happens for the first run.
        resultingFirstChild = newFiber;
      } else {
        // TODO: Defer siblings if we're not at the right index for this slot.
        // I.e. if we had null values before, then we want to defer this
        // for each null value. However, we also don't want to call updateSlot
        // with the previous one.
        previousNewFiber.sibling = newFiber;
      }
      previousNewFiber = newFiber;
      oldFiber = nextOldFiber;  // 重新给 oldFiber 赋值继续遍历
  }

// react-reconciler/src/ReactChildFiber.js line 544
function updateSlot(
    returnFiber: Fiber,
    oldFiber: Fiber | null,
    newChild: any,
    expirationTime: ExpirationTime,
  ): Fiber | null {
    // Update the fiber if the keys match, otherwise return null.

    const key = oldFiber !== null ? oldFiber.key : null;

    if (typeof newChild === 'string' || typeof newChild === 'number') {
      // 对于新的节点如果是 string 或者 number，那么都是没有 key 的，
      // 所有如果老的节点有 key 的话，就不能复用，直接返回 null。
      // 老的节点 key 为 null 的话，代表老的节点是文本节点，就可以复用
      if (key !== null) {
        return null;
      }
      return updateTextNode(
        returnFiber,
        oldFiber,
        '' + newChild,
        expirationTime,
      );
    }

    // ...

    return null;
}

// react-reconciler/src/ReactChildFiber.js line 569
if (typeof newChild === 'object' && newChild !== null) {
  switch (newChild.$$typeof) {
    case REACT_ELEMENT_TYPE: { // ReactElement 
      if (newChild.key === key) {
        if (newChild.type === REACT_FRAGMENT_TYPE) {
          return updateFragment(
            returnFiber,
            oldFiber,
            newChild.props.children,
            expirationTime,
            key,
          );
        }
        return updateElement(
          returnFiber,
          oldFiber,
          newChild,
          expirationTime,
        );
      } else {
        return null;
      }
    }
    case REACT_PORTAL_TYPE: {
      // 调用 updatePortal
      // ...
    }
  }

  if (isArray(newChild) || getIteratorFn(newChild)) {
    if (key !== null) {
      return null;
    }

    return updateFragment(
      returnFiber,
      oldFiber,
      newChild,
      expirationTime,
      null,
    );
  }
}

// react-reconciler/src/ReactChildFiber.js line 839
// 新节点已经更新完成，删除多余的老节点
if (newIdx === newChildren.length) {
  // We've reached the end of the new children. We can delete the rest.
  deleteRemainingChildren(returnFiber, oldFiber);
  return resultingFirstChild;
}

// 新节点已经更新完成，删除多余的老节点
if (oldFiber === null) {
  // If we don't have any more existing children we can choose a fast path
  // since the rest will all be insertions.
  for (; newIdx < newChildren.length; newIdx++) {
    const newFiber = createChild(
      returnFiber,
      newChildren[newIdx],
      expirationTime,
    );
    if (newFiber === null) {
      continue;
    }
    lastPlacedIndex = placeChild(newFiber, lastPlacedIndex, newIdx);
    if (previousNewFiber === null) {
      // TODO: Move out of the loop. This only happens for the first run.
      resultingFirstChild = newFiber;
    } else {
      previousNewFiber.sibling = newFiber;
    }
    previousNewFiber = newFiber;
  }
  return resultingFirstChild;
}

// react-reconciler/src/ReactChildFiber.js line 872
// Add all children to a key map for quick lookups.
// 从oldFiber开始将已经存在的节点的key或者index添加到map结构中
const existingChildren = mapRemainingChildren(returnFiber, oldFiber);

// Keep scanning and use the map to restore deleted items as moves.
// 剩余没有对比的新节点，到旧节点的map中通过key或者index一一对比查看是否可以复用。
for (; newIdx < newChildren.length; newIdx++) {
  // 主要查看新旧节点的key或者index是否有相同的，然后再查看是否可以复用。
  const newFiber = updateFromMap(
    existingChildren,
    returnFiber,
    newIdx,
    newChildren[newIdx],
    expirationTime,
  );
  if (newFiber !== null) {
    if (shouldTrackSideEffects) {
      if (newFiber.alternate !== null) {
        // The new fiber is a work in progress, but if there exists a
        // current, that means that we reused the fiber. We need to delete
        // it from the child list so that we don't add it to the deletion
        // list.
        existingChildren.delete(  // 在map中删除掉已经复用的节点的key或者index
          newFiber.key === null ? newIdx : newFiber.key,
        );
      }
    }
    lastPlacedIndex = placeChild(newFiber, lastPlacedIndex, newIdx);
    // 添加newFiber到更新过的newFiber结构中。
    if (previousNewFiber === null) {
      resultingFirstChild = newFiber;
    } else {
      previousNewFiber.sibling = newFiber;
    }
    previousNewFiber = newFiber;
  }
}

// react-reconciler/src/ReactChildFiber.js line 299
// 将旧节点的key或者index，旧节点保存到map结构中，方便通过key或者index获取
function mapRemainingChildren(
    returnFiber: Fiber,
    currentFirstChild: Fiber,
  ): Map<string | number, Fiber> {
    // Add the remaining children to a temporary map so that we can find them by
    // keys quickly. Implicit (null) keys get added to this set with their index
    // instead.
    const existingChildren: Map<string | number, Fiber> = new Map();

    let existingChild = currentFirstChild;
    while (existingChild !== null) {
      if (existingChild.key !== null) {
        existingChildren.set(existingChild.key, existingChild);
      } else {
        existingChildren.set(existingChild.index, existingChild);
      }
      existingChild = existingChild.sibling;
    }
    return existingChildren;
  }