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fix: fix potential memory leak in BTCCache #160

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merged 3 commits into from
Mar 8, 2023
Merged

fix: fix potential memory leak in BTCCache #160

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f1e1ddf
init
SebastianElvis Mar 7, 2023
dfc9be5
fix
SebastianElvis Mar 8, 2023
2a45218
minor
SebastianElvis Mar 8, 2023
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12 changes: 11 additions & 1 deletion types/btccache.go
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Original file line number Diff line number Diff line change
Expand Up @@ -59,6 +59,9 @@ func (b *BTCCache) Add(ib *IndexedBlock) {
// Thread-unsafe version of Add
func (b *BTCCache) add(ib *IndexedBlock) {
if b.size() >= b.maxEntries {
// dereference the 0-th block to ensure it will be garbage-collected
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Given that the above checks b.size() >= b.maxEntries, there's a case in which we need to remove more than only one element to have b.maxEntries elements. I know this case can't appear in the current code, but since we have such condition, we should handle it properly. Otherwise, if we're sure that the case b.size() > b.maxEntries can't happen, we should panic in this case and handle the below using a condition b.size() == b.maxEntries

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Agree, I'm inclined to panic when b.size() > b.maxEntries since BTC cache does not support adding multiple elements at the same time

// see https://stackoverflow.com/questions/55045402/memory-leak-in-golang-slice
b.blocks[0] = nil
b.blocks = b.blocks[1:]
}

Expand All @@ -85,6 +88,8 @@ func (b *BTCCache) RemoveLast() error {
return ErrEmptyCache
}

// dereference the last block to ensure it will be garbage-collected
b.blocks[len(b.blocks)-1] = nil
b.blocks = b.blocks[:len(b.blocks)-1]
return nil
}
Expand All @@ -94,7 +99,7 @@ func (b *BTCCache) RemoveAll() {
b.Lock()
defer b.Unlock()

b.blocks = b.blocks[:0]
b.blocks = []*IndexedBlock{}
}

// Size returns the size of the cache. Thread-safe.
Expand Down Expand Up @@ -210,5 +215,10 @@ func (b *BTCCache) Trim() {
return
}

// dereference b.blocks[:len(b.blocks)-int(b.maxEntries)] to ensure they will be garbage-collected
for i := range b.blocks[:len(b.blocks)-int(b.maxEntries)] {
b.blocks[i] = nil
}

b.blocks = b.blocks[len(b.blocks)-int(b.maxEntries):]
}