python always create a file name "c", why ? (with cs=imp.load_source('cs','',f) function)

Question

I would like to use a txt file with configure variable for my python code.And I saw and copied following code from web to my python.

def getVarFromFile(filename):
  import imp 
  f = open(filename)
  global cs
  cs = imp.load_source('cs', '', f)
  f.close()

The strange thing is whenever I run the python code, I got a file named "c" in my directory, and it stores lots of strange characters like:

mò ;€-Rc

Why there comes this "c" file ? How can I get rid of that ? Thanks

----adding info----- I made the call by

getVarFromFile('config/cs.txt')

And my cs.txt looks like:

var1 = 'abc'
var2 = 'h'
var3 = '2'
var4 = '1'

And I use the data like:

var2 = int(cs.var2)

And by my debugging, once I put the

sys.exit()

Before

cs = imp.load_source('cs', '', f)

There is no such c file is generated. While I put exit after that line, there is the file named "c" generated.

So it must relate to that function

Answer 1

You need to give a more sensible value for the pathname parameter.

imp.load_source(name, pathname, file) is used to load a sourcefile. As part of that it will compile the sourcefile to bytecode and write the compiled code out to a .pyc file. The name of the .pyc file I think it gets by putting c on the end of the pathname you supplied.

If you don't want the compile caching, then simply read the file and create your own module. See Load module from string in python for a way to do this:

import sys,imp

my_code = 'a = 5'
mymodule = imp.new_module('mymodule')
exec my_code in mymodule.__dict__    

Combined with your original code would be something like (untested):

def getVarFromFile(filename):
  import imp 
  with open(filename) as f:
      global cs
      cs = imp.new_module('cs')
      exec(f.read(), cs.__dict__)