Continuing from the question mentioned in the below link
list/array of sockets in python
Is it possible to create an array of sockets in python like
list=[socket0, socket1, socket2, socket3 ]
for i in range(0,3):
list[i]=socket.socket(socket.AF_INET, socket.SOCK_STREAM)
list[i].connect((host,port[0]))
I tried this but I'm getting the same error as I have posted in the link that no attribute recv.
Is it possible to create an array like this in python. I know it's possible in C.
You should not pre-populate your list, but create it on the fly.
There are two way how you can do that:
The "better" way:
sockets = [socket.socket(socket.AF_INET, socket.SOCK_STREAM) for _ in range(3)]
for sock in sockets:
sock.connect((host, port[0]))
The inferior way:
sockets = []
for i in range(3):
sock = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
sock.connect((host, port[0]))
sockets.append(sock)
Despite the extra iteration, the first one is better because it uses one of Pythons "best" festures for constructing a list and is shorter and more readable. The extra iteration's timing requirements is low to non-existent.
However, there is nothing which is really against the second one, so if you prefer it although it is longer, use that.
connect call. It is a network IO operation and I suspect it is going to last enough so we can say that the other code took a negligible amount of time to execute no matter how many loops are there. So it's all about readability.
The fastest way:
import socket
sockets = list(map(lambda x: x.connect(('127.0.0.1', 80)), [socket.socket(socket.AF_INET, socket.SOCK_STREAM) for _ in range(3)]))
List comprehension works faster than creation of loop with continuous append or list[i], map function does too.
More beautiful way(slightly less efficient):
import socket
not_connected_sockets = [socket.socket(socket.AF_INET, socket.SOCK_STREAM) for _ in
range(3)]
sockets = list(map(lambda x: x.connect(('127.0.0.1',
80)), not_connected_sockets))
list[i]=socket.socket...?