From c8cab398fd6d0ba59e66e944c6460ab8e051ea79 Mon Sep 17 00:00:00 2001 From: Ben Newman Date: Thu, 28 Sep 2023 11:55:37 -0400 Subject: [PATCH] Conserve total number of sorted arrays retained by canonicalStringify. --- src/utilities/common/canonicalStringify.ts | 56 ++++++++++++++++------ 1 file changed, 41 insertions(+), 15 deletions(-) diff --git a/src/utilities/common/canonicalStringify.ts b/src/utilities/common/canonicalStringify.ts index 9fa634eecc0..964b3063356 100644 --- a/src/utilities/common/canonicalStringify.ts +++ b/src/utilities/common/canonicalStringify.ts @@ -42,7 +42,7 @@ function stableObjectReplacer(key: string, value: any) { // means their prototype is either Object.prototype or null. if (proto === Object.prototype || proto === null) { const keys = Object.keys(value); - const sortedKeys = sortKeys(keys); + const sortedKeys = lookupSortedKeys(keys, true); if (sortedKeys !== keys) { const sorted = Object.create(null); // Reassigning the keys in sorted order will cause JSON.stringify to @@ -64,13 +64,19 @@ interface SortingTrie { const sortingTrieRoot: SortingTrie = { sorted: [], + // Using Object.create(null) is actually important here, since we could + // theoretically encounter strings like "__proto__" or "hasOwnProperty", which + // would be problematic if Object.prototype is in the prototype chain. next: Object.create(null), }; -// Sort the given keys using a lookup trie, and return the same (===) array in -// case it was already sorted, so we can avoid always creating a new object in -// the replacer function above. -function sortKeys(keys: readonly string[]): readonly string[] { +// Sort the given keys using a lookup trie, with an option to return the same +// (===) array in case it was already sorted, so we can avoid always creating a +// new object in the replacer function above. +function lookupSortedKeys( + keys: readonly string[], + returnKeysIfAlreadySorted: boolean, +): readonly string[] { let node = sortingTrieRoot; let alreadySorted = true; for (let k = 0, len = keys.length; k < len; ++k) { @@ -78,17 +84,37 @@ function sortKeys(keys: readonly string[]): readonly string[] { if (k > 0 && keys[k - 1] > key) { alreadySorted = false; } - const next = node.next; - node = next[key] || (next[key] = { next: Object.create(null) }); + node = node.next[key] || ( + node.next[key] = { next: Object.create(null) } + ); } + if (alreadySorted) { - // There may already be a node.sorted array that's equivalent to the - // already-sorted keys array, but if keys was already sorted, we always want - // to return that array, not node.sorted. - return node.sorted ? keys : (node.sorted = keys); + return node.sorted + // There may already be a node.sorted array that's equivalent to the + // already-sorted keys array, but if keys was already sorted, we want to + // return the keys reference as-is when returnKeysIfAlreadySorted is true. + // This behavior helps us decide whether we need to create a new object in + // the stableObjectReplacer function above. + ? (returnKeysIfAlreadySorted ? keys : node.sorted) + : (node.sorted = keys); } - // The .slice(0) is necessary so that we do not modify the original keys array - // by calling keys.sort(), and also so that we always return a new (!==) - // sorted array when keys was not already sorted. - return node.sorted || (node.sorted = keys.slice(0).sort()); + + // To conserve the total number of sorted arrays we store in the trie, we + // always use the same sorted array reference for a given set of strings, + // regardless of which permutation of the strings led to this SortingTrie + // node. To obtain this one true array, we do a little extra work to look up + // the sorted array associated with the sorted permutation, since there will + // be one unique path through the trie for the sorted permutation (even if + // there were duplicate keys). We can reuse the lookupSortedKeys function to + // perform this lookup, but we pass false for returnKeysIfAlreadySorted so it + // will return the existing array (if any) rather than the new sorted array we + // use to perform the lookup. If there is no existing array associated with + // the sorted permutation, the new array produced by keys.slice(0).sort() will + // be stored as the one true array and returned here. Since we are passing in + // an array that is definitely already sorted, this call to lookupSortedKeys + // will never actually have to call .sort(), so this lookup is always linear. + return node.sorted || ( + node.sorted = lookupSortedKeys(keys.slice(0).sort(), false) + ); }