问题
Why doesn't send()
in winsock
guarantee delivery of the all bytes you request?
This is TCP and it's blocking sockets.
Similarly, this happens when non-blocking. How can you guarantee that you send everything?
I've noticed recv()
does the same.
回答1:
If it didn't send everything, just call send
again on the rest. If blocking, you can do it immediately. If non-blocking, you can either wait or use a socket discovery method (like select
or I/O completion ports). The same goes for recv
. If you didn't get all you wanted, call recv
again. This is one of the reasons both recv
and send
return the number of bytes sent or received.
The number of bytes you pass to send
or recv
is just a limit. It can send less than that (though, unless non-blocking, usually won't). But it can definitely receive less than that. (The OS has no control over how much data it receives or when it receives it.)
TCP is implemented for you. But if you have an application protocol that involves application-level messages, then the application has to implement them. It won't happen by magic. TCP doesn't "glue the bytes together" into a message for you. TCP is a byte-stream protocol, not a message protocol. If you want messages, you have to implement them.
回答2:
This behaviour is "by design".
You can use an outer loop as shown in this example:
int sendBuffer (SOCKET ClientSocket, const char *buf, int len, int flags)
{
int num_left = len;
int num_sent;
int err = 0;
const char *cp = buf;
while (num_left > 0)
{
num_sent = send(ClientSocket, cp, num_left, flags);
if (num_sent < 0)
{
err = SOCKET_ERROR;
break;
}
assert(num_sent <= num_left);
num_left -= num_sent;
cp += num_sent;
}
return (err == SOCKET_ERROR ? SOCKET_ERROR : len);
}
回答3:
send
tells you what it was able to send via its return value.
Loop until send
has cumulatively sent all the data or returns an error.
来源:https://stackoverflow.com/questions/14399691/send-not-deliver-all-bytes