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[Commandline version] Shorter --version output #3750

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@rubyFeedback

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@rubyFeedback

Hey guys,

Not sure if this gets through or not but I will at the least explain the point I am trying to make here.

Right now if you do:

scons --version

You get this output:

SCons by Steven Knight et al.:
        SCons: v4.0.1.c289977f8b34786ab6c334311e232886da7e8df1, 2020-07-17 01:50:03, by bdbaddog on ProDog2020
        SCons path: ['/usr/lib/python3.8/site-packages/SCons-4.0.1-py3.8.egg/SCons']
Copyright (c) 2001 - 2020 The SCons Foundation

The string I, as a user, am interested here is the "4.0.1" part only. This follows from this URL:

https://downloads.sourceforge.net/scons/scons-4.0.1.tar.gz

There is actually a reason I am filing this, because I have another project that tries to
find out which programs are installed; and also the versions. So I have to parse the
output.

Now, I can add a special rule for scons just fine; I grab the second line
and scan for the part between "v" and "c" and I guess this will work fine for
subsequent releases.

But at the same time I wonder why this output is so long?

To be fair, other programs can also be long, the GPL based programs typically
also mention the licence and copyright.

But compare this to gnuplot:

gnuplot --version

Output is:

gnuplot 5.4 patchlevel 0

Quite easy to parse.

Or, even better, python:

python --version

Output is:

Python 3.8.4

Anyway, if you guys want to retain the current output that is fine but I propose to
then also add something like --short-version or -V or well ... perhaps --verbose
to show the above verbose output, and instead use a much shorter output.

It's no big deal either way but I just wanted to mention it. :)

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