Merge pull request prometheus#14658 from krajorama/zero-rate-nh-quantile

Native histograms: define behavior when rate is null.

promql/promqltest/testdata/histograms.test

@@ -482,3 +482,29 @@ load_with_nhcb 5m
 eval_fail instant at 50m histogram_quantile(0.99, {__name__=~"request_duration_seconds\\d*_bucket"})
 
 eval_fail instant at 50m histogram_quantile(0.99, {__name__=~"request_duration_seconds\\d*"})
+
+# Histogram with constant buckets.
+load_with_nhcb 1m
+    const_histogram_bucket{le="0.0"} 1 1 1 1 1
+    const_histogram_bucket{le="1.0"} 1 1 1 1 1
+    const_histogram_bucket{le="2.0"} 1 1 1 1 1
+    const_histogram_bucket{le="+Inf"} 1 1 1 1 1
+
+# There is no change to the bucket count over time, thus rate is 0 in each bucket.
+eval instant at 5m rate(const_histogram_bucket[5m])
+    {le="0.0"} 0
+    {le="1.0"} 0
+    {le="2.0"} 0
+    {le="+Inf"} 0
+
+# Native histograms do not represent empty buckets, so here the zeros are implicit.
+eval instant at 5m rate(const_histogram[5m])
+    {} {{schema:-53 sum:0 count:0 custom_values:[0.0 1.0 2.0]}}
+
+# Zero buckets mean no observations, so there is no value that observations fall below,
+# which means that any quantile is a NaN.
+eval instant at 5m histogram_quantile(1.0, sum by (le) (rate(const_histogram_bucket[5m])))
+    {} NaN
+
+eval instant at 5m histogram_quantile(1.0, sum(rate(const_histogram[5m])))
+    {} NaN

promql/promqltest/testdata/native_histograms.test

@@ -784,3 +784,47 @@ eval_warn instant at 1m rate(some_metric[30s])
 # Start with exponential, end with custom.
 eval_warn instant at 30s rate(some_metric[30s])
     # Should produce no results.
+
+# Histogram with constant buckets.
+load 1m
+    const_histogram {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}} {{schema:0 sum:1 count:1 buckets:[1 1 1]}}
+
+# There is no change to the bucket count over time, thus rate is 0 in each bucket.
+# However native histograms do not represent empty buckets, so here the zeros are implicit.
+eval instant at 5m rate(const_histogram[5m])
+    {} {{schema:0 sum:0 count:0}}
+
+# Zero buckets mean no observations, thus the denominator in the average is 0
+# leading to 0/0, which is NaN.
+eval instant at 5m histogram_avg(rate(const_histogram[5m]))
+    {} NaN
+
+# Zero buckets mean no observations, so count is 0.
+eval instant at 5m histogram_count(rate(const_histogram[5m]))
+    {} 0.0
+
+# Zero buckets mean no observations and empty histogram has a sum of 0 by definition.
+eval instant at 5m histogram_sum(rate(const_histogram[5m]))
+    {} 0.0
+
+# Zero buckets mean no observations, thus the denominator in the fraction is 0,
+# leading to 0/0, which is NaN.
+eval instant at 5m histogram_fraction(0.0, 1.0, rate(const_histogram[5m]))
+    {} NaN
+
+# Workaround to calculate the observation count corresponding to NaN fraction.
+eval instant at 5m histogram_count(rate(const_histogram[5m])) == 0.0 or histogram_fraction(0.0, 1.0, rate(const_histogram[5m])) * histogram_count(rate(const_histogram[5m]))
+    {} 0.0
+
+# Zero buckets mean no observations, so there is no value that observations fall below,
+# which means that any quantile is a NaN.
+eval instant at 5m histogram_quantile(1.0, rate(const_histogram[5m]))
+    {} NaN
+
+# Zero buckets mean no observations, so there is no standard deviation.
+eval instant at 5m histogram_stddev(rate(const_histogram[5m]))
+    {} NaN
+
+# Zero buckets mean no observations, so there is no standard variance.
+eval instant at 5m histogram_stdvar(rate(const_histogram[5m]))
+    {} NaN