New to Python: my impression v. Perl/Ruby

[Wayne Folta]

I've been a long-time Perl programmer, though I've not used a boatload 
of packages nor much of the tacky OO.

A couple of years ago, I decided to look into Python and Ruby. Python 
looked OK, but not that different. I did like the indent-as-group idea, 
which was different. Ruby looked very cool. But it was impossible to 
get good documentation. It seemed like a Japanese cult with a few 
western initiates.

Well, MacOS X ships with Perl, Python, and Ruby (and PHP, and ...) so I 
recently figured I'd try them again. I still find Ruby more intriguing, 
but I've settled on Python. Why?

Well, Perl is an expedient measure that (for me) doesn't scale up and 
has that horrible OO syntax. (So I never used it much.) If you're going 
to read someone else's Perl, you had better be "in the zone".

A couple of months ago, I'd written a quick Perl script that would 
"unstick" a co-worker's POP email. The problem is the Windows-based POP 
server is too stupid to realize a message does not end in CRLF, so it 
just appends a period then CRLF thinking it's doing well. But the 
period ends up not being alone on a line, so it's not a valid 
termination to the message. So their email program hangs waiting for a 
proper termination. Sigh.

A week ago, the script broke because I hadn't properly accounted for 
the fact that the user might have hundreds of messages queued up. (I 
hadn't used a Perl POP package, which might've handled it, but just 
threw it together myself. Yes, that would've made the Perl code more 
competitive with the other two solutions.)

So I decided this might be a nice exercise to try Python. And it went 
together very quickly using Python's POP3 package. Then I decided to 
try it in Ruby. I think one little portion of the code shows the 
difference between the two cultures.

The problem is that the message is not properly terminated, so you need 
to time out and catch that timeout to realize, "hmmm, malformed". In 
Python I had:

M = poplib.POP3 ('172.16.30.1') ;
M.user ('foo') ;
M.pass_ ('bar')

num_mesgs = len (M.list ()[1])
bad_mesgs = 0

for msg in range (num_mesgs):
	try:
		M.retr (msg + 1)
... blah blah...

How do I get it to time out after 5 seconds and how do I catch that? 
The online docs are pretty good, but I had to guess that POP3 was built 
on the socket package. Looking at socket's docs I found the proper 
command and exception. I had to include:

socket.setdefaulttimeout (5.0)

before the POP3 commands to set the default socket timeout to 5 
seconds. And

except socket.timeout:

is the proper way to catch the timeout. Both of these things are in the 
socket documentation.

Contrast this with Ruby. The Ruby docs are less complete, but they did 
mention that POP was subclassed from protocol and you'd have to look at 
protocol's source to see how it works. Looking through protocol, I 
figured out what to do and it was more elegant.

The protocol class had a read_timeout, but since Ruby's mantra might be 
said to be "Real OO", the POP code had been written such that you could 
say

pop.read_timeout = 5

after the POP open and it set the timeout for that pop connection to 5 
seconds. Almost as if POP passed the read_timeout upstream to socket 
automatically. (I don't think Ruby does, but it's coded to look that 
way.)

My experience is limited, but it feels like this example gives a good 
feel for the two languages. Python was better documented and things 
came together quickly. Ruby ultimately had a more elegant solution, but 
was more poorly documented. This echoes the mantras: Python's is 
"Batteries Included", while Ruby's might be "Real OO".

On a personal note, I usually prefer an elegant solution, but when the 
language goes so far as to consider

0.step(360, 45)

to be reasonable, my head hurts. I know there's syntactic sugar so I 
don't have to code like that. I know everything's an object. But, 
dammit, a constant integer is an integer with constant value and 
causing it to iterate is a step too far.
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