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trie: reduce allocations in stacktrie #30743
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Ah wait 24B that is the size of a slice: a pointer and two ints. It is the slice passed by value, escaped to heap (correctly so). But how to avoid it?? |
Solved!
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goos: linux goarch: amd64 pkg: github.com/ethereum/go-ethereum/trie cpu: 12th Gen Intel(R) Core(TM) i7-1270P │ stacktrie.3 │ stacktrie.4 │ │ sec/op │ sec/op vs base │ Insert100K-8 69.50m ± 12% 74.59m ± 14% ~ (p=0.128 n=7) │ stacktrie.3 │ stacktrie.4 │ │ B/op │ B/op vs base │ Insert100K-8 4.640Mi ± 0% 3.112Mi ± 0% -32.93% (p=0.001 n=7) │ stacktrie.3 │ stacktrie.4 │ │ allocs/op │ allocs/op vs base │ Insert100K-8 226.7k ± 0% 126.7k ± 0% -44.11% (p=0.001 n=7)
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Did a sync on |
@holiman I think we should compare the relevant metrics (memory allocation, etc) during the snap sync and to see the overall impact brought by this change. |
@@ -40,6 +40,20 @@ func (n *fullNode) encode(w rlp.EncoderBuffer) { | |||
w.ListEnd(offset) | |||
} | |||
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func (n *fullnodeEncoder) encode(w rlp.EncoderBuffer) { |
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Can you estimate the performance and allocation differences between using this encoder and the standard full node encoder?
The primary distinction seems to be that this encoder inlines the children encoding directly, rather than invoking the children encoder recursively. I’m curious about the performance implications of this approach.
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Sure, so if I change it back (adding back rawNode, and then switching back the case branchNode:
so that it looks like it did earlier, then the difference is:
[user@work go-ethereum]$ benchstat pr.bench pr.bench..2
goos: linux
goarch: amd64
pkg: github.com/ethereum/go-ethereum/trie
cpu: 12th Gen Intel(R) Core(TM) i7-1270P
│ pr.bench │ pr.bench..2 │
│ sec/op │ sec/op vs base │
Insert100K-8 88.67m ± 33% 88.20m ± 11% ~ (p=0.579 n=10)
│ pr.bench │ pr.bench..2 │
│ B/op │ B/op vs base │
Insert100K-8 3.451Ki ± 3% 2977.631Ki ± 0% +86178.83% (p=0.000 n=10)
│ pr.bench │ pr.bench..2 │
│ allocs/op │ allocs/op vs base │
Insert100K-8 22.00 ± 5% 126706.50 ± 0% +575838.64% (p=0.000 n=10)
When we build the children struct, all the values are copied, as opposed if we just use the encoder-type which just uses the same child without triggering a copy.
n := shortNode{Key: hexToCompactInPlace(st.key), Val: valueNode(st.val)} | ||
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n.encode(t.h.encbuf) | ||
{ |
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We can also use shortNodeEncoder here?
I would recommend not to do the inline encoding directly.
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We could, but here we know we have a valueNode
. And the valueNode just does:
func (n valueNode) encode(w rlp.EncoderBuffer) {
w.WriteBytes(n)
}
If we change that to a method which checks the size and theoretically does different things, then the semantics become slightly changed.
//shortNodeEncoder is a type used exclusively for encoding. Briefly instantiating | ||
// a shortNodeEncoder and initializing with existing slices is less memory | ||
// intense than using the shortNode type. | ||
shortNodeEncoder struct { |
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qq, does shortNodeEncoder
need to implement the node interface?
It's just an encoder which coverts the node object to byte format?
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just have a try, it's technically possible
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@@ -27,6 +27,7 @@ import ( | |||
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var ( | |||
stPool = sync.Pool{New: func() any { return new(stNode) }} | |||
bPool = newBytesPool(32, 100) |
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any particular reason to not use sync.Pool
?
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Yes! You'd be surprised, but the whole problem I had with extra alloc :
I solved that by not using a sync.Pool
. I suspect it's due to the interface-conversion, but I don't know any deeper details about the why of it.
@@ -129,6 +143,12 @@ const ( | |||
) | |||
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func (n *stNode) reset() *stNode { | |||
if n.typ == hashedNode { |
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What's the difference between HashedNode and Embedded tiny node?
For embedded node, we also allocate the buffer from the bPool and the buffer is owned by the node itself right?
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Well, the thing is, that there's only one place in the stacktrie where we convert a node into a hashedNode
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st.typ = hashedNode
st.key = st.key[:0]
st.val = nil // Release reference to potentially externally held slice.
This is the only place. So we know that once something has been turned into a hashedNode, the val
is never externally held and thus it can be returned to the pool. The big problem we had earlier, is that we need to overwrite the val
with a hash, but we are not allowed to mutate val
. So this was a big cause of runaway allocs.
But now we reclaim those values-which-are-hashes since we know that they're "ours".
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For embedded node, we also allocate the buffer from the bPool and the buffer is owned by the node itself right?
To answer your question: yes. So from the perspective of slice-reuse, there is zero difference.
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Btw, in my follow-up PR to reduce allocs in derivesha:#30747 , I remove this trick again.
In that PR, I always copy the value as it enters the stacktrie. So we always own the val
slice, and are free to reuse via pooling.
Doing so is less hacky, we get rid of this special case "val is externally held unless it's an hashedNode because then we own it".
That makes it possible for the derivesha-method to reuse the input-buffer
This PR uses various tweaks and tricks to make the stacktrie near alloc-free.
Also improves derivesha: