pip.parse(python_version=) should not be mandatory #1708
Description
🐞 bug report
Affected Rule
Bzlmod pip.parse()
Is this a regression?
Yes, was not a problem in WORKSPACE.
Description
The arg documentation says the following:
https://github.com/bazelbuild/rules_python/blob/711186f144af06b431bd416b2d742874de3a2dea/python/private/bzlmod/pip.bzl#L391-L398
It specifically describes If not specified, then the default Python version (as set by the root module or rules_python) will be used. So this attribute should be optional but currently is mandatory.
🔬 Minimal Reproduction
pip.parse() without python_version.
🔥 Exception or Error
ERROR: in tag at <root>/MODULE.bazel:42:10, mandatory attribute python_version isn't being specified
🌍 Your Environment
Operating System:
Linux
Output of bazel version:
2024/01/19 15:28:42 Warning: used fallback version "6.3.2"
Bazelisk version: v1.19.0
INFO: Running bazel wrapper (see //tools/bazel for details), bazel version 6.1.2 will be used instead of system-wide bazel installation.
Build label: 6.1.2
Build target: bazel-out/k8-opt/bin/src/main/java/com/google/devtools/build/lib/bazel/BazelServer_deploy.jar
Build time: Tue Apr 18 15:29:54 2023 (1681831794)
Build timestamp: 1681831794
Build timestamp as int: 1681831794
Rules_python version:
0.27.1
Anything else relevant?