-
Notifications
You must be signed in to change notification settings - Fork 3
New issue
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
How to get the name from an index? #12
Comments
Hi!
|
I noticed that there are faults in JuliaIO/ZipArchives.jl#89 on a 32-bit system with Ubuntu and on Windows. It seems the issue is with the structure:
The field types need to be correctly defined for different systems. |
Yes, not just this struct; the other functions need to be fixed to match https://github.com/nih-at/libzip/blob/main/lib/zip.h Lines 384 to 386 in bfa63f4
For example based on: https://github.com/nih-at/libzip/blob/e9083a7b41463489cd6641310c36f3b9197e1d27/lib/zip.h#L499 function libzip_source_stat(source, sb)
return ccall((:zip_source_stat, libzip), Cint, (Ptr{LibZipSourceT}, Ref{LibZipStatT}), source, sb)
end You may want to check out https://github.com/JuliaInterop/Clang.jl to help with this. |
If I have
ZipArchive
and anindex
, how can I get the entry name as aString
?I am currently using:
But this seems to be causing segmentation faults in JuliaIO/ZipArchives.jl#89
Is there a less unsafe way to do this?
To clarify, I want the raw bytes of the entry name, I don't want any "automagical" encoding conversions.
The text was updated successfully, but these errors were encountered: