regexp_replace fails when pattern or replacement is a scalar NULL #11410
Description
Describe the bug
regexp_replace fails to produce correct number of rows if either pattern or replacement arg is a scalar NULL.
I think this is due to
fetch_string_arg not passing the correct length to the _regexp_replace_early_abort function - when this specific arg is a scalar, its "array len" is just 1, and the abort function creates a 1-len array as the result.
To Reproduce
Normal case - values are valid scalars or array nulls:
> select regexp_replace(col, 'a', 'c') from (values ('a'), ('b')) as tbl(col);
+---------------------------------------------+
| regexp_replace(tbl.col,Utf8("a"),Utf8("c")) |
+---------------------------------------------+
| c |
| b |
+---------------------------------------------+
2 row(s) fetched.
Elapsed 0.001 seconds.
> select regexp_replace(col, ncol, 'c') from (values ('a', NULL), ('b', NULL)) as tbl(col, ncol);
+--------------------------------------------+
| regexp_replace(tbl.col,tbl.ncol,Utf8("c")) |
+--------------------------------------------+
| |
| |
+--------------------------------------------+
Failing case - pattern or replacement is a scalar NULL:
> select regexp_replace(col, NULL, 'c') from (values ('a'), ('b')) as tbl(col);
Internal error: UDF returned a different number of rows than expected. Expected: 2, Got: 1.
This was likely caused by a bug in DataFusion's code and we would welcome that you file an bug report in our issue tracker
> select regexp_replace(col, 'a', NULL) from (values ('a'), ('b')) as tbl(col);
Internal error: UDF returned a different number of rows than expected. Expected: 2, Got: 1.
This was likely caused by a bug in DataFusion's code and we would welcome that you file an bug report in our issue tracker
Expected behavior
No response
Additional context
No response