Remove sharding for VecCache
This sharding is never used (per the comment in code). If we re-add sharding at some point in the future this is cheap to restore, but for now no need for the extra complexity.
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@ -101,7 +101,7 @@ fn iter(&self, f: &mut dyn FnMut(&Self::Key, &Self::Value, DepNodeIndex)) {
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}
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pub struct VecCache<K: Idx, V> {
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cache: Sharded<IndexVec<K, Option<(V, DepNodeIndex)>>>,
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cache: Lock<IndexVec<K, Option<(V, DepNodeIndex)>>>,
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}
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impl<K: Idx, V> Default for VecCache<K, V> {
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@ -120,24 +120,20 @@ impl<K, V> QueryCache for VecCache<K, V>
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#[inline(always)]
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fn lookup(&self, key: &K) -> Option<(V, DepNodeIndex)> {
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// FIXME: lock_shard_by_hash will use high bits which are usually zero in the index() passed
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// here. This makes sharding essentially useless, always selecting the zero'th shard.
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let lock = self.cache.lock_shard_by_hash(key.index() as u64);
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let lock = self.cache.lock();
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if let Some(Some(value)) = lock.get(*key) { Some(*value) } else { None }
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}
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#[inline]
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fn complete(&self, key: K, value: V, index: DepNodeIndex) {
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let mut lock = self.cache.lock_shard_by_hash(key.index() as u64);
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let mut lock = self.cache.lock();
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lock.insert(key, (value, index));
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}
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fn iter(&self, f: &mut dyn FnMut(&Self::Key, &Self::Value, DepNodeIndex)) {
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for shard in self.cache.lock_shards() {
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for (k, v) in shard.iter_enumerated() {
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if let Some(v) = v {
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f(&k, &v.0, v.1);
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}
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for (k, v) in self.cache.lock().iter_enumerated() {
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if let Some(v) = v {
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f(&k, &v.0, v.1);
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}
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}
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}
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@ -149,9 +145,6 @@ pub struct DefIdCache<V> {
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///
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/// The second element of the tuple is the set of keys actually present in the IndexVec, used
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/// for faster iteration in `iter()`.
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// FIXME: This may want to be sharded, like VecCache. However *how* to shard an IndexVec isn't
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// super clear; VecCache is effectively not sharded today (see FIXME there). For now just omit
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// that complexity here.
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local: Lock<(IndexVec<DefIndex, Option<(V, DepNodeIndex)>>, Vec<DefIndex>)>,
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foreign: DefaultCache<DefId, V>,
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}
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