Mark Johnston <ma...@freebsd.org> wrote: > There is an additional factor: wasted space. When writing data to a > socket, the kernel buffers that data in mbufs. All mbufs have some > amount of embedded storage, and the kernel accounts for that storage, > whether or not it's used. With small byte datagrams there can be a lot > of overhead;
I'm observing two mbufs being allocated for each datagram for small datagrams, but only one mbuf for larger datagrams. That seems counter-intuitive to me? > The kern.ipc.sockbuf_waste_factor sysctl controls the upper limit on > total bytes (used or not) that may be enqueued in a socket buffer. The > default value of 8 means that we'll waste up to 7 bytes per byte of > data, I think. Setting it higher should let you enqueue more messages. Ah, this looks like something relevant. Setting kern.ipc.sockbuf_waste_factor=1, I can only write 8 1-byte datagrams. For any increase of the waste factor by one, I get another 8 1-byte datagrams, up until waste factor > 29, at which point we hit recvspace: 30 * 8 = 240, so 240 1-byte datagrams with 16 bytes dgram overhead means we get 240*17 = 4080 bytes, which just fits (well, with room for one empty 16-byte dgram) into the recvspace = 4096. But I still don't get the direct relationship between the waste factor and the recvspace / buffer queue: with a waste_factor of 1 and a datagram with 1972 bytes, I'm able to write one dgram with 1972 bytes + 1 dgram with 1520 bytes = 3492 bytes (plus 2 * 16 bytes overhead = 3524 bytes). There'd still have been space for 572 more bytes in the second dgram. Liekwise, trying to write a single 1973 dgram fills the queue and no additional bytes can be written in a second dgram, but I can write a single 2048 byte dgram. Still confused... -Jan
