using if statement with list comparison

Question

Here is my problem: on reading a book on networking programming for python, i stumbles across this code:

import socket, sys
s = socket.socket(socket.AF_INET, socket.SOCK_DGRAM)

if sys.argv[1:] == ['server']:
    s.bind('127.0.0.1', PORT)
    ...
    ...

and so on. My question is whether the if statement checks if any of the elements in the sys.argv list(except the first item) is compared to be equal to 'server'. I tried doing this in IDLE for Python 3.2 and it didn't work. The book is intended for python 2.7 so I tried that too but it still dint work.

score 3 · Answer 1 · edited May 23 '17 at 12:29

3

No, it checks whether the list formed by the slice 1: is equal to the list ['server'].

edited May 23 '17 at 12:29

Community

1
1

answered Aug 29 '13 at 13:36

mikołak

9,455
1
46
69

score 3 · Answer 2 · answered Aug 29 '13 at 13:37

3

sys.argv[0] is the name of the program, which is stripped. sys.argv[1:] - all the command line arguments provided to the program.

The if statement checks that your script received only one argument server.

answered Aug 29 '13 at 13:37

mishik

9,873
9
42
65

score 3 · Answer 3 · answered Aug 29 '13 at 13:37

3

No, that wouldn't work in any version of Python. The only thing that does is to check that sys.argv from position 1 onwards has only one element, which is 'server'.

answered Aug 29 '13 at 13:37

Daniel Roseman

567,968
59
825
842

score 2 · Answer 4 · answered Aug 29 '13 at 14:01

2

You can check 'server' argument whith this construction:

if 'server' in sys.argv[1:]:
    do_something()

For future, use argparse for get command line arguments.

answered Aug 29 '13 at 14:01

Michael Kazarian

4,166
1
21
24

using if statement with list comparison

4 Answers4