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fix(lru): don't clear an already cached key on set #725

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fix(lru): don't clear an already cached key on set #725

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4 changes: 2 additions & 2 deletions internal/lrucache.js
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Original file line number Diff line number Diff line change
Expand Up @@ -21,9 +21,9 @@ class LRUCache {
}

set (key, value) {
const deleted = this.delete(key)
this.delete(key)

Comment on lines +24 to 25
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Suggested change
this.delete(key)

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@jviide jviide Jul 13, 2024 •
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Removing the this.delete(key) would turn the LRU cache implementation into a mixture of LRU and FIFO caches. It may not be relevant with the module's current usage pattern. However, it could lead to a surprise down the road if the cache module's usage is expanded.

I suggest either keeping the delete call, removing the delete while calling the cache implementation something else than an LRU, or just closing this pull request as unnecessary (as it does not affect currently observable behavior like you pointed out). I ran the benchmarks and removing or keeping the delete call seems to make no difference there, though there are currently no benchmarks that heavily exercise the .set method.

if (!deleted && value !== undefined) {
if (value !== undefined) {
// If cache is full, delete the least recently used item
if (this.map.size >= this.max) {
const firstKey = this.map.keys().next().value
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22 changes: 22 additions & 0 deletions test/internal/lrucache.js
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Original file line number Diff line number Diff line change
Expand Up @@ -17,3 +17,25 @@ test('basic cache operation', t => {
c.set(42, undefined)
t.end()
})

test('test setting the same key twice', t => {
const c = new LRUCache()
c.set(1, 1)
c.set(1, 2)
t.equal(c.get(1), 2)
t.end()
})

test('test that setting an already cached key bumps it last in the LRU queue', t => {
const c = new LRUCache()
const max = 1000

for (let i = 0; i < max; i++) {
c.set(i, i)
}
c.set(0, 0)
c.set(1001, 1001)
t.equal(c.get(0), 0)
t.equal(c.get(1), undefined)
t.end()
})
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