React源码分析之diff核心算法

React的diff算法是在render的beginWork阶段中进行处理

beginWork是在向下深度遍历fiber树时会对途径的每个节点进行状态处理和进行diff对比

首先diff的入口是在reconcileChildFibers中，然后会根据type来判断使用哪种diff函数进行处理

function reconcileChildFibers(
  returnFiber: Fiber,
  currentFirstChild: Fiber | null,
  newChild: any,
  lanes: Lanes,
): Fiber | null {
  if (typeof newChild === 'object' && newChild !== null) {
    switch (newChild.$$typeof) {
      case REACT_ELEMENT_TYPE:
        return placeSingleChild(
          reconcileSingleElement(
            returnFiber,
            currentFirstChild,
            newChild,
            lanes,
          ),
        );··
      case REACT_PORTAL_TYPE:
        // ...
      case REACT_LAZY_TYPE:
        //...
    }

    if (isArray(newChild)) {
      return reconcileChildrenArray(
        returnFiber,
        currentFirstChild,
        newChild,
        lanes,
      );
    }

    if (getIteratorFn(newChild)) {
      //...
    }
  }
  // ...
}

我在本篇会针对两种较常用的diff函数进行分析

reconcileSingleElement是针对新newChild是单节点，而oldChild单节点或者是多节点就无法确定了，所以在此diff算法中就会对旧节点进行遍历，然后删除不匹配的oldFiber

function reconcileSingleElement(
  returnFiber: Fiber,
  currentFirstChild: Fiber | null,
  element: ReactElement
  lanes: Lanes,
): Fiber {
  const key = element.key;
  let child = currentFirstChild;
  /**
    * 遍历旧节点，找到与newChild相同key的节点，不匹配的删除
    * 针对匹配的oldFiber, 用newChild中新节点的props来生成新的fiber节点
    */
  while (child !== null) {
    if (child.key === key) {
      const elementType = element.type;
      /**
        * 通过useFiber创建一个新的Fiber
        * 如果element是一个Fragment，则以element.props.children建立Fiber
        * 将returnFiber赋给新的fiber的return字段，然后返回这个新的fiber
        */·
      if (elementType === REACT_FRAGMENT_TYPE) {
        if (child.tag === Fragment) {
          deleteRemainingChildren(returnFiber, child.sibling);
          const existing = useFiber(child, element.props.children);
          existing.return = returnFiber;
          if (__DEV__) {
            existing._debugSource = element._source;
            existing._debugOwner = element._owner;
          }
          return existing;
        }
      } else {
        if (
          child.elementType === elementType ||
          (__DEV__
            ? isCompatibleFamilyForHotReloading(child, element)
            : false) ||
          (typeof elementType === 'object' &&
            elementType !== null &&
            elementType.$$typeof === REACT_LAZY_TYPE &&
            resolveLazy(elementType) === child.type)
        ) {
          deleteRemainingChildren(returnFiber, child.sibling);
          const existing = useFiber(child, element.props);
          existing.ref = coerceRef(returnFiber, child, element);
          existing.return = returnFiber;
          if (__DEV__) {
            existing._debugSource = element._source;
            existing._debugOwner = element._owner;
          }
          return existing;
        }
      }
      // Didn't match.
      deleteRemainingChildren(returnFiber, child);
      break;
    } else {
      // key不相同就删除
      deleteChild(returnFiber, child);
    }
    child = child.sibling;
  }

  // 如果没有命中一个key，则通过createFiberFormElement或CreateFiberFormFragment创建一个新的fiber，然后返回
  if (element.type === REACT_FRAGMENT_TYPE) {
    const created = createFiberFromFragment(
      element.props.children,
      returnFiber.mode,
      lanes,
      element.key,
    );
    created.return = returnFiber;
    return created;
  } else {
    const created = createFiberFromElement(element, returnFiber.mode, lanes);
    created.ref = coerceRef(returnFiber, currentFirstChild, element);
    created.return = returnFiber;
    return created;
  }
}

针对newChild是多节点的情况就需要调用reconcileChildrenArray进行diff操作

多节点会有四种可能性的变化：删除、新增、位移、更新

reconcileChildrenArray针对这四种变化，首先会处理的是更新，当出现无法匹配的情况时，就会根据遍历的情况来判断是否处理删除或者新增，然后最后会根据情况处理位移

因为fiber是单向链表，所以reconcileChildrenArray的遍历不是双端遍历

首先第一轮遍历，是处理节点更新

for (; oldFiber !== null && newIdx < newChildren.length; newIdx++) {
  // newChildren遍历完了，oldFiber没有遍历完，中断遍历
  if (oldFiber.index > newIdx) {
    nextOldFiber = oldFiber;
    oldFiber = null;
  } else {
    // 记录oldFiber的下一个节点
    nextOldFiber = oldFiber.sibling;
  }
  // 更新节点，如果节点没有匹配上，就会返回null
  const newFiber = updateSlot(
    returnFiber,
    oldFiber,
    newChildren[newIdx],
    lanes,
  );
  // newFiber为null说明节点没有匹配上，中断遍历
  if (newFiber === null) {
    // oldFiber为null说明oldFiber也遍历完了
    if (oldFiber === null) {
      oldFiber = nextOldFiber;
    }
    break;
  }

  /**
   * shouldTrackSideEffects为true表示是更新过程
   * mountChildFibers = ChildReconciler(false);
   * reconcileChildFibers = ChildReconciler(true);
   * ChildReconciler接收的就是shouldTrackSideEffects
   */
  if (shouldTrackSideEffects) {
    if (oldFiber && newFiber.alternate === null) {
      // 新节点没有现有节点，需要删除
      deleteChild(returnFiber, oldFiber);
    }
  }
  // 记录固定节点的位置
  lastPlacedIndex = placeChild(newFiber, lastPlacedIndex, newIdx);

  // 将新节点拼接成以sibling为指针的单向链表
  if (previousNewFiber === null) {
    resultingFirstChild = newFiber;
  } else {
    previousNewFiber.sibling = newFiber;
  }
  previousNewFiber = newFiber;
  oldFiber = nextOldFiber;
}

遍历完匹配的节点后，就判断新节点是否遍历完，如果遍历完，那么剩余的oldFiber都是要删除的

if (newIdx === newChildren.length) {
  deleteRemainingChildren(returnFiber, oldFiber);
  if (getIsHydrating()) {
    const numberOfForks = newIdx;
    pushTreeFork(returnFiber, numberOfForks);
  }
  return resultingFirstChild;
}

如果新旧点没有遍历完，就判断旧fiber链是否遍历完，如果遍历完那么剩余的新节点全部作为新fiber插入

if (oldFiber === null) {
  for (; newIdx < newChildren.length; newIdx++) {
    // 创建新fiber节点
    const newFiber = createChild(returnFiber, newChildren[newIdx], lanes);
    if (newFiber === null) {
      continue;
    }

    // 记录固定节点
    lastPlacedIndex = placeChild(newFiber, lastPlacedIndex, newIdx);

    // 将新fiber拼接成以sibling为指针的单向链表
    if (previousNewFiber === null) {
      resultingFirstChild = newFiber;
    } else {
      previousNewFiber.sibling = newFiber;
    }
    previousNewFiber = newFiber;
  }
  if (getIsHydrating()) {
    const numberOfForks = newIdx;
    pushTreeFork(returnFiber, numberOfForks);
  }
  return resultingFirstChild;
}

执行到这一步，说明新旧节点都没有遍历完，就说明存在有位移的未知序列

// 首先创建一个以oldFiber key为键，值为oldFiber的map
const existingChildren = mapRemainingChildren(returnFiber, oldFiber);

for (; newIdx < newChildren.length; newIdx++) {
  // 然后根据map中的oldFiber创建新fiber
  const newFiber = updateFromMap(
    existingChildren,
    returnFiber,
    newIdx,
    newChildren[newIdx],
    lanes,
  );
  if (newFiber !== null) {
    if (shouldTrackSideEffects) {
      if (newFiber.alternate !== null) {
        // 如果newFiber.alternate不为null，说明是根据oldFiber创建的，那么就需要在map中删除oldFiber
        existingChildren.delete(
          newFiber.key === null ? newIdx : newFiber.key,
        );
      }
    }

    // 根据lastPlacedIndex判断是否移动节点
    lastPlacedIndex = placeChild(newFiber, lastPlacedIndex, newIdx);

    // 将新fiber拼接成以sibling为指针的单向链表
    if (previousNewFiber === null) {
      resultingFirstChild = newFiber;
    } else {
      previousNewFiber.sibling = newFiber;
    }
    previousNewFiber = newFiber;
  }
}

if (shouldTrackSideEffects) {
  // 删除剩余的oldFiber
  existingChildren.forEach(child => deleteChild(returnFiber, child));
}

移动节点的核心是在placeChild这个函数中，如果当前正在遍历的节点的oldIndex是在lastPlacedIndex的右边，就说明它的位置没变化，因为旧节点中就处于右边，新节点中也处于右边。

function placeChild(
  newFiber: Fiber,
  lastPlacedIndex: number,
  newIndex: number,
): number {
  newFiber.index = newIndex;
  if (!shouldTrackSideEffects) {
    newFiber.flags |= Forked;
    return lastPlacedIndex;
  }
  const current = newFiber.alternate;
  if (current !== null) {
    const oldIndex = current.index;
    if (oldIndex < lastPlacedIndex) {
      // 小于lastPlacedIndex 产生了位移
      newFiber.flags |= Placement | PlacementDEV;
      return lastPlacedIndex;
    } else {
      // 没有位移，返回当前的oldIndex
      return oldIndex;
    }
  } else {
    newFiber.flags |= Placement | PlacementDEV;
    return lastPlacedIndex;
  }
}

针对单节点的diff，会遍历oldFiber链，如果有匹配的fiber，就以匹配的生成新fiber，如果没有就新建一个fiber，然后删除不匹配的fiber

针对多节点diff