LSSTApplications  10.0+286,10.0+36,10.0+46,10.0-2-g4f67435,10.1+152,10.1+37,11.0,11.0+1,11.0-1-g47edd16,11.0-1-g60db491,11.0-1-g7418c06,11.0-2-g04d2804,11.0-2-g68503cd,11.0-2-g818369d,11.0-2-gb8b8ce7
LSSTDataManagementBasePackage
Using lsstDebug to control debugging output

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:

import lsstDebug
print lsstDebug.Info(__name__).display

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

import lsstDebug
def DebugInfo(name):
di = lsstDebug.getInfo(name) # N.b. lsstDebug.Info(name) would call us recursively
if name == "foo":
di.display = True
return di
lsstDebug.Info = DebugInfo

into a file debug.py and

import lsstDebug
print "display is", lsstDebug.Info(__name__).display

into foo.py, then

$ python -c "import foo"
display is False

but

$ python -c "import debug; import foo"
display is True

The command line task interface supports a flag –debug to import debug.py from your PYTHONPATH