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Revert "Merge pull request #1399 from aweary/use-react-reconciler-rea…
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Brandon Dail
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packages/enzyme-adapter-react-16/src/findCurrentFiberUsingSlowPath.js
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// Extracted from https://github.com/facebook/react/blob/7bdf93b17a35a5d8fcf0ceae0bf48ed5e6b16688/src/renderers/shared/fiber/ReactFiberTreeReflection.js#L104-L228 | ||
function findCurrentFiberUsingSlowPath(fiber) { | ||
const { alternate } = fiber; | ||
if (!alternate) { | ||
return fiber; | ||
} | ||
// If we have two possible branches, we'll walk backwards up to the root | ||
// to see what path the root points to. On the way we may hit one of the | ||
// special cases and we'll deal with them. | ||
let a = fiber; | ||
let b = alternate; | ||
while (true) { // eslint-disable-line | ||
const parentA = a.return; | ||
const parentB = parentA ? parentA.alternate : null; | ||
if (!parentA || !parentB) { | ||
// We're at the root. | ||
break; | ||
} | ||
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// If both copies of the parent fiber point to the same child, we can | ||
// assume that the child is current. This happens when we bailout on low | ||
// priority: the bailed out fiber's child reuses the current child. | ||
if (parentA.child === parentB.child) { | ||
let { child } = parentA; | ||
while (child) { | ||
if (child === a) { | ||
// We've determined that A is the current branch. | ||
return fiber; | ||
} | ||
if (child === b) { | ||
// We've determined that B is the current branch. | ||
return alternate; | ||
} | ||
child = child.sibling; | ||
} | ||
// We should never have an alternate for any mounting node. So the only | ||
// way this could possibly happen is if this was unmounted, if at all. | ||
throw new Error('Unable to find node on an unmounted component.'); | ||
} | ||
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if (a.return !== b.return) { | ||
// The return pointer of A and the return pointer of B point to different | ||
// fibers. We assume that return pointers never criss-cross, so A must | ||
// belong to the child set of A.return, and B must belong to the child | ||
// set of B.return. | ||
a = parentA; | ||
b = parentB; | ||
} else { | ||
// The return pointers point to the same fiber. We'll have to use the | ||
// default, slow path: scan the child sets of each parent alternate to see | ||
// which child belongs to which set. | ||
// | ||
// Search parent A's child set | ||
let didFindChild = false; | ||
let { child } = parentA; | ||
while (child) { | ||
if (child === a) { | ||
didFindChild = true; | ||
a = parentA; | ||
b = parentB; | ||
break; | ||
} | ||
if (child === b) { | ||
didFindChild = true; | ||
b = parentA; | ||
a = parentB; | ||
break; | ||
} | ||
child = child.sibling; | ||
} | ||
if (!didFindChild) { | ||
// Search parent B's child set | ||
({ child } = parentB); | ||
while (child) { | ||
if (child === a) { | ||
didFindChild = true; | ||
a = parentB; | ||
b = parentA; | ||
break; | ||
} | ||
if (child === b) { | ||
didFindChild = true; | ||
b = parentB; | ||
a = parentA; | ||
break; | ||
} | ||
child = child.sibling; | ||
} | ||
if (!didFindChild) { | ||
throw new Error('Child was not found in either parent set. This indicates a bug ' + | ||
'in React related to the return pointer. Please file an issue.'); | ||
} | ||
} | ||
} | ||
} | ||
if (a.stateNode.current === a) { | ||
// We've determined that A is the current branch. | ||
return fiber; | ||
} | ||
// Otherwise B has to be current branch. | ||
return alternate; | ||
} | ||
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module.exports = findCurrentFiberUsingSlowPath; |