P1673: "whose type is a template parameter with `Out` in its name" is outdated
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Description
https://github.com/ORNL/cpp-proposals-pub/blob/d8f530fa7698597ea99e37e593f7d88628812962/D1673/P1673.md?plain=1#L5630 ([linalg.general]) speak of "a function parameter whose type is a template parameter with Out in its name." We have exposition-only concepts now to talk about output and in/out parameters, so this language is likely outdated.
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- Read the whole issue, then the project's contributing guide.
- Comment on the issue to say you are picking it up — it saves two people doing the same work.
- Fork the repository and make your change on a branch.
- Open a pull request that references the issue number.
Research direction
Open D1673/P1673.md at the referenced [linalg.general] passage around line 5630 and read the surrounding wording. Check how the paper's exposition-only concepts describe output and in/out parameters, then update the outdated reference to a template parameter with Out in its name and verify the passage remains consistent.
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Assessment
- Tech stack
- cpp
- Domain
- documentation
- Issue type
- Documentation
- Difficulty
- 2/5
- Estimated time
- 1-3 hours
- Activity status
- Stale
- Clarity
- Mostly clear
- Newbie friendliness
- 58/100