argv listing directory files
<p>this very simple code is driving me nuts.</p>

<pre><code>import sys,string,socket  

maquina = sys.argv[1]
texto = string.join(sys.argv[2:], " ")
print '['+maquina+']' 
print '['+texto+']'

mysock = socket.socket( socket.AF_INET, socket.SOCK_DGRAM )
if  maquina == '*':
    for ip in range(18,254):
        mysock.sendto( texto, ('192.168.0.'+str(ip),8090) )
else:
    mysock.sendto( texto, (maquina,8090))
mysock.close()
</code></pre>

<p>When I type </p>

<pre><code>python send.py 192.168.0.135 several strings 
</code></pre>

<p>the output is:</p>

<pre><code>[192.168.0.135]
[several strings]
</code></pre>

<p>But if I type </p>

<pre><code>python send.py * several strings
</code></pre>

<p>I got:</p>

<pre><code>[ajax]
[ajustaprofile.py build dia2django.py djangoajax django-cube django-evolution django_excel_templates-0.1 Doc google_appengine httpd.conf imagens jre1.5.0_08 lib list_mailbackup.py luma models.py netsend.py pip-log.txt reverse.py share testparamet.py tnsnames.ora util virttool-0.1 xlrd-0.7.1 xlutils-1.4.1 xlwt-0.7.2 several lines]
</code></pre>

<p>This is the file list that I have on current directory!
Python translates somehow '*' to the files of current directory.</p>

<p>I tried to do the homework using this link but I got no clue.</p>

<p><a href="http://docs.python.org/library/sys.html" rel="nofollow">http://docs.python.org/library/sys.html</a>
<a href="http://www.faqs.org/docs/diveintopython/kgp_commandline.html" rel="nofollow">http://www.faqs.org/docs/diveintopython/kgp_commandline.html</a></p>

<p>How can I use "*" on argv ?</p>
