Howto get source line from python generator object?

Question

Here is an example:

def g():
  yield str('123')
  yield int(123)
  yield str('123')

o = g()

while True:
  v = o.next()
  if isinstance( v, str ):
    print 'Many thanks to our generator...'
  else:
    # Or GOD! I don't know what to do with this type
    #
    raise TypeError( '%s:%d Unknown yield value type %s.' % \
                     (g.__filename__(), g.__lineno__(), type(v) )
                   )

How I could get source file name and exact yield line number, when my generator returns unknown type (int in this example) ?

Answer 1

I don't think you can get the information you want. That's the sort of thing captured in exceptions, but there has been no exception here.

Answer 2

Your generator object "o" in this case has all the information you want - You can pasting your example ona python Console, and inspecting with "dir" both the function "g" and the generator "o".

The generator has the attributes "gi_code" and "gi_frame" which in turn contain the information you want:

>>> o.gi_code.co_filename
'<stdin>'
#this is the line neumber inside the file:
>>> o.gi_code.co_firstlineno
1
# and this is the current line number inside the function:
>>> o.gi_frame.f_lineno
3