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Issue #21566: Make use of socket.listen() default backlog.

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This commit is contained in:
Charles-François Natali 2014-07-23 19:28:13 +01:00
parent 8518b79a8d
commit 6e20460dc6
18 changed files with 33 additions and 33 deletions
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8
Lib/test/test_timeout.py
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@ -243,14 +243,14 @@ def testRecvTimeout(self):
def testAcceptTimeout(self):
# Test accept() timeout
support.bind_port(self.sock, self.localhost)
self.sock.listen(5)
self.sock.listen()
self._sock_operation(1, 1.5, 'accept')
def testSend(self):
# Test send() timeout
with socket.socket(socket.AF_INET, socket.SOCK_STREAM) as serv:
support.bind_port(serv, self.localhost)
serv.listen(5)
serv.listen()
self.sock.connect(serv.getsockname())
# Send a lot of data in order to bypass buffering in the TCP stack.
self._sock_operation(100, 1.5, 'send', b"X" * 200000)
@ -259,7 +259,7 @@ def testSendto(self):
# Test sendto() timeout
with socket.socket(socket.AF_INET, socket.SOCK_STREAM) as serv:
support.bind_port(serv, self.localhost)
serv.listen(5)
serv.listen()
self.sock.connect(serv.getsockname())
# The address argument is ignored since we already connected.
self._sock_operation(100, 1.5, 'sendto', b"X" * 200000,
@ -269,7 +269,7 @@ def testSendall(self):
# Test sendall() timeout
with socket.socket(socket.AF_INET, socket.SOCK_STREAM) as serv:
support.bind_port(serv, self.localhost)
serv.listen(5)
serv.listen()
self.sock.connect(serv.getsockname())
# Send a lot of data in order to bypass buffering in the TCP stack.
self._sock_operation(100, 1.5, 'sendall', b"X" * 200000)