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LSSTApplications
1.1.2+25,10.0+13,10.0+132,10.0+133,10.0+224,10.0+41,10.0+8,10.0-1-g0f53050+14,10.0-1-g4b7b172+19,10.0-1-g61a5bae+98,10.0-1-g7408a83+3,10.0-1-gc1e0f5a+19,10.0-1-gdb4482e+14,10.0-11-g3947115+2,10.0-12-g8719d8b+2,10.0-15-ga3f480f+1,10.0-2-g4f67435,10.0-2-gcb4bc6c+26,10.0-28-gf7f57a9+1,10.0-3-g1bbe32c+14,10.0-3-g5b46d21,10.0-4-g027f45f+5,10.0-4-g86f66b5+2,10.0-4-gc4fccf3+24,10.0-40-g4349866+2,10.0-5-g766159b,10.0-5-gca2295e+25,10.0-6-g462a451+1
LSSTDataManagementBasePackage
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The class lsstDebug can be used to turn on debugging output in a non-intrusive way. For example, the variable lsstDebug.Info("lsst.meas.astrom.astrom").debug is used to control debugging output from the lsst.meas.astrom.astrom module.
It is always safe to interrogate lsstDebug; for example lsstDebug.Info("Robert.Hugh.Lupton").isBadPerson will return False.
The convention is that the name ("lsst.meas.astrom.astrom") is the __name__ of the module, so the source code will typically look something like:
which will print False unless lsstDebug.Info(__name__).display has somehow been set to True.
Why is this interesting? Because you can replace lsstDebug.Info with your own version, e.g. if you put
into a file debug.py and
into foo.py, then
but
The command line task interface supports a flag –debug to import debug.py from your PYTHONPATH
1.8.5