How to determine file, function and line number?

In C++, I can print debug output like this:

printf(
   "FILE: %s, FUNC: %s, LINE: %d, LOG: %s\n",
   __FILE__,
   __FUNCTION__,
   __LINE__,
   logmessage
);

How can I do something similar in Python?

There is a module named inspect which provides these information.

Example usage:

import inspect

def PrintFrame():
  callerframerecord = inspect.stack()[1]    # 0 represents this line
                                            # 1 represents line at caller
  frame = callerframerecord[0]
  info = inspect.getframeinfo(frame)
  print(info.filename)                      # __FILE__     -> Test.py
  print(info.function)                      # __FUNCTION__ -> Main
  print(info.lineno)                        # __LINE__     -> 13

def Main():
  PrintFrame()                              # for this line

Main()

However, please remember that there is an easier way to obtain the name of the currently executing file:

print(__file__)

For example

import inspect
frame = inspect.currentframe()
# __FILE__
fileName  =  frame.f_code.co_filename
# __LINE__
fileNo = frame.f_lineno

There's more here http://docs.python.org/library/inspect.html

Building on geowar's answer:

class __LINE__(object):
    import sys

    def __repr__(self):
        try:
            raise Exception
        except:
            return str(sys.exc_info()[2].tb_frame.f_back.f_lineno)

__LINE__ = __LINE__()

If you normally want to use __LINE__ in e.g. print (or any other time an implicit str() or repr() is taken), the above will allow you to omit the ()s.

(Obvious extension to add a __call__ left as an exercise to the reader.)