python - Why does this try/except/else function return None? -

i trying create directory. if name exists, appends "_1" directory name recursively.

this runs fine if directory not exist already. runs fine if directory exists (it creates directory+"_1" , returns same). however, if directory + "_1" exists, function creates directory+"_1_1" returns none.

import os  def recurs_mkdir(out_dir):     try:         os.mkdir(out_dir)     except oserror:         out_dir += "_1"         recurs_mkdir(out_dir)     else:         print "returning: %s" % out_dir         return out_dir  >>> print recurs_mkdir("existing_folder") returning: existing_folder_1 none 

why return none in case of exception?

edit: nneoneo's answer works if "exisiting_folder" exists, not behave if "existing_folder" , "existing_folder_1" exist. modified, function is

import os  def recurs_mkdir(out_dir):     try:         os.mkdir(out_dir)     except oserror:         out_dir += "_1"         recurs_mkdir(out_dir)         print "returning: %s" % out_dir         return out_dir     else:         print "returning: %s" % out_dir         return out_dir 

if "existing_folder" created:

>>> recurs_mkdir("existing_folder") returning: existing_folder_1 returning: existing_folder_1 existing_folder_1 

if "existing_folder" , "existing_folder_1" exist:

>>> recurs_mkdir("existing_folder") returning: existing_folder_1_1 returning: existing_folder_1_1 returning: existing_folder_1 existing_folder_1 

edit 2: answer nneoneo

import os  def recurs_mkdir(out_dir):     try:         os.mkdir(out_dir)     except oserror:         out_dir += "_1"         return recurs_mkdir(out_dir)     else:         print "returning: %s" % out_dir         return out_dir 

because no return statement invoked @ in case of exception, function returns default value of none (this same return value you'd if wrote bare return statement without return value).

add return statement except clause return whatever want.