Hi guys.. i found something strange on the behavior of OptionParser

If I have this sample code : 

import sys
from optparse import OptionParser

if __name__ == '__main__':
    parser = OptionParser()
    parser.add_option("-p", "--p", help="The P of python", default=None)
    parser.add_option("-y", "--y", help="The Y of python", default=None)
        
    (options,args) = parser.parse_args(sys.argv)
    print options

and i execute :

myFile.py -p -y

the options.p will be -y and options.y will be None

worst.. if my user say :

myFile.py -p -y 'thon'

options.p will be -y and options.y will be None...

In all case I think that, if i do 

myFile.py -p -y 

options -p and options -y must be None

and if i want to put -y in options.p i say something like 

myFile.py -p "-y" so i can do after myFile.py -p "-y" -y "thon".

This is not a bug.  By default, options in optparse take an argument --
therefore, the -y is taken as the argument to the -p option.

Use e.g. add_option(..., action='store_true') to specify an option that
doesn't take an argument.