mbutrovich commented on code in PR #11710:
URL: https://github.com/apache/arrow-rs/pull/11710#discussion_r4339284213
##########
arrow-array/src/array/run_array.rs:
##########
@@ -73,12 +73,62 @@ macro_rules! ree_map {
/// └ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─ ─┘ │ B │
/// ├─────────────────┤
/// RunArray │ B │ run length
of 'B' = run_ends[2] - run_ends[1] = 3
-/// length = 3 └─────────────────┘
+/// offset = 0, len = 6 └─────────────────┘
///
/// Logical array
/// Contents
/// ```
///
+/// # `offset` and `len`
+///
+/// The [`Array::len`] and [`Array::offset`] for a `RunArray` refer to the
+/// **logical** length, not the `run_ends` and `values` child arrays. In the
+/// example above the `RunArray` has length `6` even though `run_ends` and
+/// `values` each contain `3` elements.
+///
+/// The values in `run_ends` represent indexes of the *unsliced* logical array.
+/// This allows zero-copy [`RunArray::slice`]: `run_ends` and `values` are left
+/// untouched and only `offset` and `len` are adjusted.
+///
+/// A [`RunArray`] must satisfy the following invariants:
+///
+/// * every run end is greater than zero and `run_ends` is strictly increasing
+/// * `run_ends` has no null values
+/// * `run_ends` and `values` have the same length
+/// * if `len > 0`, the last run end is greater than or equal to `offset +
len`.
Review Comment:
> if `len > 0`, the last run end is greater than or equal to `offset + len`.
Is the `len > 0` condition intended? The [REE
layout](https://github.com/apache/arrow/blob/373d64c02cb23663b63c5853099fca43a46c39fc/docs/source/format/Columnar.rst?plain=1#L1089-L1160)
does not mention a parent offset, so as I read it the spec is silent. C++
[requires the last run end to cover `offset + length` when the run ends are
non-empty](https://github.com/apache/arrow/blob/373d64c02cb23663b63c5853099fca43a46c39fc/cpp/src/arrow/util/ree_util.cc#L247-L254),
even for `length == 0`. With run ends `[1]`, `offset = 2` and `len = 0`,
`validate_full` accepts the array on `main` and, as I read the C++ code, C++
rejects it.
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