[SciPy-dev] What's this error mean?

Pearu Peterson pearu at cens.ioc.ee
Thu Sep 26 10:29:51 CDT 2002


On Thu, 26 Sep 2002, Skip Montanaro wrote:

> I just tried a 'cvs up' in my scipy source directory and got this:
> 
>     cvs [server aborted]: could not find desired version 1.130.1506.4149 in /home/cvsroot/world/scipy/__cvs_version__.py,v
> 
> I removed __cvs_version__.py and ran 'cvs up' again.  It then worked.
>
> Can someone explain what that error means and what __cvs_version__.py is
> used for?

I am not sure why you recieved this error, may be __cvs_version__.py
was modified in your side for some reasons...

__cvs_version__.py is used for holding the CVS version information of
SciPy. This version information is calculated at the CVS server side and
it uses Revision numbers of the files in the repository.
So, whenever any change is made to the CVS repository, the version
information of SciPy will be updated automatically and the algorithm
ensures that it will be always monotonically increasing, even if files
are removed (so that some Revision are lost).

This CVS version information is used also for setting the real SciPy
version. For example, the last two bits of

>>> scipy.__version__
'0.2.0_alpha_133.4175'

come from __cvs_version__.py. This makes any SciPy checkout from CVS 
unique in the sense that if an user reports problems, then from these
two number one can tell what CVS state this particular user was
using. This information can be sometimes helpful for tracking down bugs.

In addition, no brain power is wasted for setting the version string when
making a snapshot of SciPy.

Pearu




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