Fixes Iterator boundaries #2136
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The start should be inclusive and the end is supposed to be exclusive. If start and end are equal, then only one entry can be returned which need to also be equal to start. Fixes grafana#2124 Those issues were edge cases where the boundaries of a block or iterator would be equal to the start. Signed-off-by: Cyril Tovena <cyril.tovena@gmail.com>
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Codecov Report
@@ Coverage Diff @@
## master #2136 +/- ##
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+ Coverage 61.23% 61.39% +0.16%
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Files 146 146
Lines 11196 11202 +6
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+ Hits 6856 6878 +22
+ Misses 3794 3777 -17
- Partials 546 547 +1
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| if maxt > b.mint && b.maxt > mint { | ||
| its = append(its, b.iterator(ctx, c.readers, filter)) | ||
| if maxt < b.mint || b.maxt < mint { | ||
| continue |
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Just thinking out loud, clarifying how this addresses #2124.
With the patch: ignore the current block when none of its entries is within the interval [mint, maxt]. Otherwise append it for further processing. Or, in other words: if the current block has any entry within the interval [mint, maxt] then append it for further processing.
The special case discussed in #2124 corresponds to b.maxt == mint in this code section here, and with the old behavior maxt > b.mint && b.maxt > mint a block containing an entry within the interval [mint, maxt] was erroneously ignored for further processing.
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| if ok && (i.maxt.Before(ts) || i.maxt.Equal(ts)) { // The maxt is exclusive. | ||
| ok = false | ||
| if ok { |
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couldn't this more simply be expressed as
if i.maxt.Before(ts) || i.maxt.Equal(ts) {
ok = false
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No this is a very specific path covering the case where both maxt == mint == ts.
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The shortcut is required. I tried it but the test wasn't passing.
The start should be inclusive and the end is supposed to be exclusive.
If start and end are equal, then only one entry can be returned which needs to also be equal to start.
Fixes #2124
Those issues were edge cases where the boundaries of a block or iterator would be equal to the start.
Signed-off-by: Cyril Tovena cyril.tovena@gmail.com