prometheus/tsdb/ooo_head.go
Bryan Boreham e7e50a3afd TSDB: Remove code for querying OOO-head only
Just query via `HeadAndOOOQuerier`, which will skip series where no
in-order chunks are in range.

Now we don't need `OOORangeHead`.

Signed-off-by: Bryan Boreham <bjboreham@gmail.com>
2024-08-14 13:41:13 +01:00

168 lines
5.0 KiB
Go

// Copyright 2022 The Prometheus Authors
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
package tsdb
import (
"sort"
"github.com/prometheus/prometheus/model/histogram"
"github.com/prometheus/prometheus/tsdb/chunkenc"
)
// OOOChunk maintains samples in time-ascending order.
// Inserts for timestamps already seen, are dropped.
// Samples are stored uncompressed to allow easy sorting.
// Perhaps we can be more efficient later.
type OOOChunk struct {
samples []sample
}
func NewOOOChunk() *OOOChunk {
return &OOOChunk{samples: make([]sample, 0, 4)}
}
// Insert inserts the sample such that order is maintained.
// Returns false if insert was not possible due to the same timestamp already existing.
func (o *OOOChunk) Insert(t int64, v float64, h *histogram.Histogram, fh *histogram.FloatHistogram) bool {
// Although out-of-order samples can be out-of-order amongst themselves, we
// are opinionated and expect them to be usually in-order meaning we could
// try to append at the end first if the new timestamp is higher than the
// last known timestamp.
if len(o.samples) == 0 || t > o.samples[len(o.samples)-1].t {
o.samples = append(o.samples, sample{t, v, h, fh})
return true
}
// Find index of sample we should replace.
i := sort.Search(len(o.samples), func(i int) bool { return o.samples[i].t >= t })
if i >= len(o.samples) {
// none found. append it at the end
o.samples = append(o.samples, sample{t, v, h, fh})
return true
}
// Duplicate sample for timestamp is not allowed.
if o.samples[i].t == t {
return false
}
// Expand length by 1 to make room. use a zero sample, we will overwrite it anyway.
o.samples = append(o.samples, sample{})
copy(o.samples[i+1:], o.samples[i:])
o.samples[i] = sample{t, v, h, fh}
return true
}
func (o *OOOChunk) NumSamples() int {
return len(o.samples)
}
// ToEncodedChunks returns chunks with the samples in the OOOChunk.
//
//nolint:revive // unexported-return.
func (o *OOOChunk) ToEncodedChunks(mint, maxt int64) (chks []memChunk, err error) {
if len(o.samples) == 0 {
return nil, nil
}
// The most common case is that there will be a single chunk, with the same type of samples in it - this is always true for float samples.
chks = make([]memChunk, 0, 1)
var (
cmint int64
cmaxt int64
chunk chunkenc.Chunk
app chunkenc.Appender
)
prevEncoding := chunkenc.EncNone // Yes we could call the chunk for this, but this is more efficient.
for _, s := range o.samples {
if s.t < mint {
continue
}
if s.t > maxt {
break
}
encoding := chunkenc.EncXOR
if s.h != nil {
encoding = chunkenc.EncHistogram
} else if s.fh != nil {
encoding = chunkenc.EncFloatHistogram
}
// prevApp is the appender for the previous sample.
prevApp := app
if encoding != prevEncoding { // For the first sample, this will always be true as EncNone != EncXOR | EncHistogram | EncFloatHistogram
if prevEncoding != chunkenc.EncNone {
chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
}
cmint = s.t
switch encoding {
case chunkenc.EncXOR:
chunk = chunkenc.NewXORChunk()
case chunkenc.EncHistogram:
chunk = chunkenc.NewHistogramChunk()
case chunkenc.EncFloatHistogram:
chunk = chunkenc.NewFloatHistogramChunk()
default:
chunk = chunkenc.NewXORChunk()
}
app, err = chunk.Appender()
if err != nil {
return
}
}
switch encoding {
case chunkenc.EncXOR:
app.Append(s.t, s.f)
case chunkenc.EncHistogram:
// Ignoring ok is ok, since we don't want to compare to the wrong previous appender anyway.
prevHApp, _ := prevApp.(*chunkenc.HistogramAppender)
var (
newChunk chunkenc.Chunk
recoded bool
)
newChunk, recoded, app, _ = app.AppendHistogram(prevHApp, s.t, s.h, false)
if newChunk != nil { // A new chunk was allocated.
if !recoded {
chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
cmint = s.t
}
chunk = newChunk
}
case chunkenc.EncFloatHistogram:
// Ignoring ok is ok, since we don't want to compare to the wrong previous appender anyway.
prevHApp, _ := prevApp.(*chunkenc.FloatHistogramAppender)
var (
newChunk chunkenc.Chunk
recoded bool
)
newChunk, recoded, app, _ = app.AppendFloatHistogram(prevHApp, s.t, s.fh, false)
if newChunk != nil { // A new chunk was allocated.
if !recoded {
chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
cmint = s.t
}
chunk = newChunk
}
}
cmaxt = s.t
prevEncoding = encoding
}
if prevEncoding != chunkenc.EncNone {
chks = append(chks, memChunk{chunk, cmint, cmaxt, nil})
}
return chks, nil
}