/**
* Copyright (c) Meta Platforms, Inc. and affiliates.
*
* This source code is licensed under the MIT license found in the
* LICENSE file in the root directory of this source tree.
*
* @flow
*/
// Ids are base 32 strings whose binary representation corresponds to the
// position of a node in a tree.
// Every time the tree forks into multiple children, we add additional bits to
// the left of the sequence that represent the position of the child within the
// current level of children.
//
// 00101 00010001011010101
// ╰─┬─╯ ╰───────┬───────╯
// Fork 5 of 20 Parent id
//
// The leading 0s are important. In the above example, you only need 3 bits to
// represent slot 5. However, you need 5 bits to represent all the forks at
// the current level, so we must account for the empty bits at the end.
//
// For this same reason, slots are 1-indexed instead of 0-indexed. Otherwise,
// the zeroth id at a level would be indistinguishable from its parent.
//
// If a node has only one child, and does not materialize an id (i.e. does not
// contain a useId hook), then we don't need to allocate any space in the
// sequence. It's treated as a transparent indirection. For example, these two
// trees produce the same ids:
//
// <> <>
// <Indirection> <A />
// <A /> <B />
// </Indirection> </>
// <B />
// </>
//
// However, we cannot skip any node that materializes an id. Otherwise, a parent
// id that does not fork would be indistinguishable from its child id. For
// example, this tree does not fork, but the parent and child must have
// different ids.
//
// <Parent>
// <Child />
// </Parent>
//
// To handle this scenario, every time we materialize an id, we allocate a
// new level with a single slot. You can think of this as a fork with only one
// prong, or an array of children with length 1.
//
// It's possible for the size of the sequence to exceed 32 bits, the max
// size for bitwise operations. When this happens, we make more room by
// converting the right part of the id to a string and storing it in an overflow
// variable. We use a base 32 string representation, because 32 is the largest
// power of 2 that is supported by toString(). We want the base to be large so
// that the resulting ids are compact, and we want the base to be a power of 2
// because every log2(base) bits corresponds to a single character, i.e. every
// log2(32) = 5 bits. That means we can lop bits off the end 5 at a time without
// affecting the final result.
export type TreeContext = {
+id: number,
+overflow: string,
};
export const emptyTreeContext = {
id: 1,
overflow: '',
};
export function getTreeId(context: TreeContext): string {
const overflow = context.overflow;
const idWithLeadingBit = context.id;
const id = idWithLeadingBit & ~getLeadingBit(idWithLeadingBit);
return id.toString(32) + overflow;
}
export function pushTreeContext(
baseContext: TreeContext,
totalChildren: number,
index: number,
): TreeContext {
const baseIdWithLeadingBit = baseContext.id;
const baseOverflow = baseContext.overflow;
// The leftmost 1 marks the end of the sequence, non-inclusive. It's not part
// of the id; we use it to account for leading 0s.
const baseLength = getBitLength(baseIdWithLeadingBit) - 1;
const baseId = baseIdWithLeadingBit & ~(1 << baseLength);
const slot = index + 1;
const length = getBitLength(totalChildren) + baseLength;
// 30 is the max length we can store without overflowing, taking into
// consideration the leading 1 we use to mark the end of the sequence.
if (length > 30) {
// We overflowed the bitwise-safe range. Fall back to slower algorithm.
// This branch assumes the length of the base id is greater than 5; it won't
// work for smaller ids, because you need 5 bits per character.
//
// We encode the id in multiple steps: first the base id, then the
// remaining digits.
//
// Each 5 bit sequence corresponds to a single base 32 character. So for
// example, if the current id is 23 bits long, we can convert 20 of those
// bits into a string of 4 characters, with 3 bits left over.
//
// First calculate how many bits in the base id represent a complete
// sequence of characters.
const numberOfOverflowBits = baseLength - (baseLength % 5);
// Then create a bitmask that selects only those bits.
const newOverflowBits = (1 << numberOfOverflowBits) - 1;
// Select the bits, and convert them to a base 32 string.
const newOverflow = (baseId & newOverflowBits).toString(32);
// Now we can remove those bits from the base id.
const restOfBaseId = baseId >> numberOfOverflowBits;
const restOfBaseLength = baseLength - numberOfOverflowBits;
// Finally, encode the rest of the bits using the normal algorithm. Because
// we made more room, this time it won't overflow.
const restOfLength = getBitLength(totalChildren) + restOfBaseLength;
const restOfNewBits = slot << restOfBaseLength;
const id = restOfNewBits | restOfBaseId;
const overflow = newOverflow + baseOverflow;
return {
id: (1 << restOfLength) | id,
overflow,
};
} else {
// Normal path
const newBits = slot << baseLength;
const id = newBits | baseId;
const overflow = baseOverflow;
return {
id: (1 << length) | id,
overflow,
};
}
}
function getBitLength(number: number): number {
return 32 - clz32(number);
}
function getLeadingBit(id: number) {
return 1 << (getBitLength(id) - 1);
}
// TODO: Math.clz32 is supported in Node 12+. Maybe we can drop the fallback.
const clz32 = Math.clz32 ? Math.clz32 : clz32Fallback;
// Count leading zeros.
// Based on:
// https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Math/clz32
const log = Math.log;
const LN2 = Math.LN2;
function clz32Fallback(x: number): number {
const asUint = x >>> 0;
if (asUint === 0) {
return 32;
}
return (31 - ((log(asUint) / LN2) | 0)) | 0;
}