New versions are merging old and new file contents about zbox HOT 2 CLOSED

@gameldar thank you very much for this issue.

This is an expected result. What happened is that the 2nd short file write was overwriting the file on top of previous version.

For example, suppose we have a long buffer [1, 2, 3, 4, 5] and a short buffer [6, 7, 8]. If we write the long buffer we got the 1st version of content:

version 1: [1, 2, 3, 4, 5]

Then we write the short buffer, we are actually writing on top of the version 1 file from the beginning, not the original empty file, so we got:

version 2: [6, 7, 8, 4, 5]

That's what we're expecting. It is exactly the same behavior if you use std::fs to write a file.

fn main() {
    let buf = [1, 2, 3, 4, 5];
    let buf2 = [6, 7, 8];

    let mut f = std::fs::OpenOptions::new()
        .write(true)
        .create(true)
        .open("tt.bin")
        .unwrap();
    f.write_all(&buf).unwrap();

    let mut f = std::fs::OpenOptions::new()
        .write(true)
        .create(true)
        .open("tt.bin")
        .unwrap();
    f.write_all(&buf2).unwrap();
 }

The above code will produce the file:

$ hexdump ./tt.bin
0000000 06 07 08 04 05

If you want to write a fresh content to the file, you can consider add truncate option, which will clear the content before writing.

// at the 2nd short file write
let mut f = OpenOptions::new()
        .create(true)
        .truncate(true)
        .open(&mut repo, "/myfile.txt")?;

from zbox.

Ahh yes I see now. Most of the time in the past (in other languages) I've inherently passed through the truncate flag without realising it (e.g. it is the default behaviour).

But the implementation here is exactly what I want to use in another use case as well and using the truncate solves my immediate issue.

Thanks for the quick reply!

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