Re: How to interrogate optical discs

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On Tue, Jun 15, 2010 at 6:33 PM, Richard Shaw <[email protected]> wrote:
> I did some searching but found no definitive answer. One thing that
> could improve the image size estimation is to round up the size of
> each file to a multiple of 2k. This still will not take into account
> file system metadata overhead but it will get us closer and should be
> pretty easy to implement (if it's not already, I'll check the code).

Replying to myself here. I'm not sure what's going on but I forced the
size of each file to round up to the nearest 2K block but I got a
strange result. I have it spit out the cumulative file size of each
disc and for disc one I got 4679MB, which would be fine in MiB but not
MB, but the resulting ISO file was 4476MB.

Here's part of the code: (line numbers added)
<CODE>
1     if not os.path.islink(file):
2        file_size = os.path.getsize(file)
3        iso_size = math.ceil(file_size/(2*1024))*2*1024
4        disc_size += iso_size
</CODE>

You don't have to know python to interpret it so please let me know if
you see anything wrong with the logic.
1. Test to make sure it's a real file and not a link to a file.
2. Get the file size in bytes.
3. Round up the file size to the nearest 2K block.
4. Add the iso_size of the file to the disc_size variable
... wash, rinse, repeat...

Any ideas?

Richard
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