do not understand this use of sys module
Question: Here is code from a tutorial in A Byte of Python:

    
    
    import sys
    
    filename = 'poem.txt'
    
    def readfile(filename):
        #Print a file to standard output
        f = file(filename)
        while True:
            line = f.readline()
            if len(line) == 0:
                break
            print line,
        f.close()
    
    if len(sys.argv) < 2:
        print 'No action specified'
        sys.exit() //<--This is where the error is occurring
    
    if sys.argv[1].startswith('--'):
        option = sys.argv[1][2:] #fetches sys.argv[1] without first 2 char
        if option == 'version':
            print 'Version 1.2'
    
        elif option == 'help':
            print '''\
    This program prints files to the standard output.
    Any number of files can be specified.
    Options include:
        --version: Prints the version number
        --help: Displays this help'''
    
        else:
            print 'Unknown option'
        sys.exit()
    
    else:
        for filename in sys.argv[1:]:
            readfile(filename)
    

When I run this code, this is the error that appears:

    
    
    Traceback (most recent call last):
      File "C:/Python/sysmodulepr.py", line 17, in <module>
        sys.exit()
    SystemExit
    

I don't understand why. Please help.


Answer: It's telling you that sys.exit() has executed on line 17 of your program.

[The entry for for sys.exit in the Python
documentation](http://docs.python.org/library/sys.html#sys.exit) tells you
that this exits your program.

There's no way this line can execute without producing other output, so I
think there's something missing in the question.

