|
| 1 | +// Extracted from https://github.com/facebook/react/blob/7bdf93b17a35a5d8fcf0ceae0bf48ed5e6b16688/src/renderers/shared/fiber/ReactFiberTreeReflection.js#L104-L228 |
| 2 | +function findCurrentFiberUsingSlowPath(fiber) { |
| 3 | + const { alternate } = fiber; |
| 4 | + if (!alternate) { |
| 5 | + return fiber; |
| 6 | + } |
| 7 | + // If we have two possible branches, we'll walk backwards up to the root |
| 8 | + // to see what path the root points to. On the way we may hit one of the |
| 9 | + // special cases and we'll deal with them. |
| 10 | + let a = fiber; |
| 11 | + let b = alternate; |
| 12 | + while (true) { // eslint-disable-line |
| 13 | + const parentA = a.return; |
| 14 | + const parentB = parentA ? parentA.alternate : null; |
| 15 | + if (!parentA || !parentB) { |
| 16 | + // We're at the root. |
| 17 | + break; |
| 18 | + } |
| 19 | + |
| 20 | + // If both copies of the parent fiber point to the same child, we can |
| 21 | + // assume that the child is current. This happens when we bailout on low |
| 22 | + // priority: the bailed out fiber's child reuses the current child. |
| 23 | + if (parentA.child === parentB.child) { |
| 24 | + let { child } = parentA; |
| 25 | + while (child) { |
| 26 | + if (child === a) { |
| 27 | + // We've determined that A is the current branch. |
| 28 | + return fiber; |
| 29 | + } |
| 30 | + if (child === b) { |
| 31 | + // We've determined that B is the current branch. |
| 32 | + return alternate; |
| 33 | + } |
| 34 | + child = child.sibling; |
| 35 | + } |
| 36 | + // We should never have an alternate for any mounting node. So the only |
| 37 | + // way this could possibly happen is if this was unmounted, if at all. |
| 38 | + throw new Error('Unable to find node on an unmounted component.'); |
| 39 | + } |
| 40 | + |
| 41 | + if (a.return !== b.return) { |
| 42 | + // The return pointer of A and the return pointer of B point to different |
| 43 | + // fibers. We assume that return pointers never criss-cross, so A must |
| 44 | + // belong to the child set of A.return, and B must belong to the child |
| 45 | + // set of B.return. |
| 46 | + a = parentA; |
| 47 | + b = parentB; |
| 48 | + } else { |
| 49 | + // The return pointers point to the same fiber. We'll have to use the |
| 50 | + // default, slow path: scan the child sets of each parent alternate to see |
| 51 | + // which child belongs to which set. |
| 52 | + // |
| 53 | + // Search parent A's child set |
| 54 | + let didFindChild = false; |
| 55 | + let { child } = parentA; |
| 56 | + while (child) { |
| 57 | + if (child === a) { |
| 58 | + didFindChild = true; |
| 59 | + a = parentA; |
| 60 | + b = parentB; |
| 61 | + break; |
| 62 | + } |
| 63 | + if (child === b) { |
| 64 | + didFindChild = true; |
| 65 | + b = parentA; |
| 66 | + a = parentB; |
| 67 | + break; |
| 68 | + } |
| 69 | + child = child.sibling; |
| 70 | + } |
| 71 | + if (!didFindChild) { |
| 72 | + // Search parent B's child set |
| 73 | + ({ child } = parentB); |
| 74 | + while (child) { |
| 75 | + if (child === a) { |
| 76 | + didFindChild = true; |
| 77 | + a = parentB; |
| 78 | + b = parentA; |
| 79 | + break; |
| 80 | + } |
| 81 | + if (child === b) { |
| 82 | + didFindChild = true; |
| 83 | + b = parentB; |
| 84 | + a = parentA; |
| 85 | + break; |
| 86 | + } |
| 87 | + child = child.sibling; |
| 88 | + } |
| 89 | + if (!didFindChild) { |
| 90 | + throw new Error('Child was not found in either parent set. This indicates a bug ' |
| 91 | + + 'in React related to the return pointer. Please file an issue.'); |
| 92 | + } |
| 93 | + } |
| 94 | + } |
| 95 | + } |
| 96 | + if (a.stateNode.current === a) { |
| 97 | + // We've determined that A is the current branch. |
| 98 | + return fiber; |
| 99 | + } |
| 100 | + // Otherwise B has to be current branch. |
| 101 | + return alternate; |
| 102 | +} |
| 103 | + |
| 104 | +module.exports = findCurrentFiberUsingSlowPath; |
0 commit comments