Fix empty filename case #204
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Codecov ReportAttention: Patch coverage is
Additional details and impacted files@@ Coverage Diff @@
## master #204 +/- ##
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- Coverage 62.37% 62.36% -0.02%
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Files 32 32
Lines 5292 5301 +9
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+ Hits 3301 3306 +5
- Misses 1991 1995 +4 ☔ View full report in Codecov by Sentry. |
| let filename = if let Some(cl_filename) = cluster.file_name() { | ||
| cl_filename.to_str().unwrap() | ||
| } else { | ||
| cluster.to_str().unwrap() |
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why should we use full path here?
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If the filename is empty it means that it ends with '.' or is '/' and we couldn't canonicalize it. Then we have no other options besides using this path as is.
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If the filename is empty it means that it ends with '.' or is '/' and we couldn't canonicalize it. Then we have no other options besides using this path as is.
Could you provide some use cases, when and why this situation happens?
It seems to me, that this code filter directories.
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The case is when casr-cli is called with an argument of directory name, and that directory contains raw casreps. For example, if casreps are located in root directory (a poked-out case, but still), or in cwd '.' and we couldn't canonicalize it for some reason. The bug was detected when calling casr-cli on cwd with raw reports.
This code adds directory to clusters if it contains raw reports.
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Can we add a comment in code stating this?
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