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LSSTApplications
11.0-13-gbb96280,12.1.rc1,12.1.rc1+1,12.1.rc1+2,12.1.rc1+5,12.1.rc1+8,12.1.rc1-1-g06d7636+1,12.1.rc1-1-g253890b+5,12.1.rc1-1-g3d31b68+7,12.1.rc1-1-g3db6b75+1,12.1.rc1-1-g5c1385a+3,12.1.rc1-1-g83b2247,12.1.rc1-1-g90cb4cf+6,12.1.rc1-1-g91da24b+3,12.1.rc1-2-g3521f8a,12.1.rc1-2-g39433dd+4,12.1.rc1-2-g486411b+2,12.1.rc1-2-g4c2be76,12.1.rc1-2-gc9c0491,12.1.rc1-2-gda2cd4f+6,12.1.rc1-3-g3391c73+2,12.1.rc1-3-g8c1bd6c+1,12.1.rc1-3-gcf4b6cb+2,12.1.rc1-4-g057223e+1,12.1.rc1-4-g19ed13b+2,12.1.rc1-4-g30492a7
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