whats this mean?

[Bill Bell]

thedustbustr at aol.com (TheDustbustr) wrote, in part:
> # BEGIN CODE BLOCK
> try:
>     s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
>     s.bind((HOST, PORT))
>     s.listen(1)
>     conn, addr = s.accept()
> except socket.error, why: ## reference 1
>     print "Server Error: " + errno.errorcode[why[0]] + ", " + why[1]
>     ##
> reference 2
>     sys.exit(1)
> # END CODE BLOCK
> 
> what does reference 2 mean?  what do those brackets do?  And for
> reference 1, what will 'why' hold if an error occurs?

Dustin,

Consider the following code that is similar to what you included in 
your message.

    from errno import *
    import socket

    try:
        raise socket.error ( 1, "an error message" )
    except socket.error, why:
        print why
        print why [ 0 ]
        print why [ 1 ]
        print "Error: " + errno.errorcode[why[0]] + ", " + why[1]

I have replaced the sockets code with a 'raise' statement that does 
what the sockets code would do in case of an exception.

When I executed this code I got the following output:

(1, 'an error')
1
an error
Error: EPERM, an error

ie, one line for each 'print' statement in the exception suite.

1. 'why' will contain a tuple whose elements are the arguments 
passed to the procedure socket.error
2. 'why[0]' is the first element in the tuple and thus is the first 
argument to socket.error; likewise 'why[1]' will contain the second 
argument.
3. 'errno.errorcode' supplies the os-specific code (ie the code that 
is meaningful for the os) corresponding to the value of the first 
argument to 'socket.error'

A pair of brackets indexes into the expression to the left of the pair. 
For example, '[0]' in 'why[0]' returns the 0th item in 'why'.

How well does this clarify the code?

Bill