[suggestion] Check that `.parse()` is called.
Nobody has claimed this yet.
Assessment
- Difficulty
- 5/5
- Estimated time
- Over a week
- Newbie friendliness
- 35/100
Research direction
Locate the parser class, its destructor, and the parse() entry point, then review existing tests for parser lifetime and exception behavior. Done means an unparsed parser raises logic_error with the requested message when destroyed, while a parser that has called parse() does not.
Written by the indexing model from the issue text.
Description
a small convenience feature: it is easy to forget calling .parse(). It would help if the destructor of the parser checks if the function was called and throws a logic_error("You forgot to call .parse()") if it was not the case.
Note, that you explicitly need to mark a destructor as noexcept(false) if you want to throw from it.
- Dominant language
- C++
- Stars
- 11
- Forks
- 9
- Avg merge
- 2h 33m
- Merged PRs (30d)
- 1
Contributor guide
First steps
- 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.
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