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From: torek@elf.ee.lbl.gov (Chris Torek)
Newsgroups: comp.unix.programmer
Subject: Re: readv/writev
Message-ID: <14609@dog.ee.lbl.gov>
Date: 24 Jun 91 09:53:58 GMT
Article-I.D.: dog.14609
References: <1991Jun21.215941.5693@ncsu.edu>
Reply-To: torek@elf.ee.lbl.gov (Chris Torek)
Organization: Lawrence Berkeley Laboratory, Berkeley
Lines: 40
X-Local-Date: Mon, 24 Jun 91 02:53:58 PDT

In article <1991Jun21.215941.5693@ncsu.edu> jwb@cepmax.ncsu.edu writes:
>Stevens (in Unix Network Programming) notes that read/write on a
>socket may input/output fewer bytes than requested.  Can I expect the
>same behavior for readv/writev?

Yes.

>If so, how might one implement "readvn"/"writevn"? (a la Steven's ...)

(I am guessing what these are: loops around the read/write calls.)
It is ugly.  You must make use of the fact that the if the system
call returns `early', it has partially processed at most one vector
(and fully processed only any earlier vectors).  Thus, something like:

	cc = readv(fd, iov, iovcnt);
	if (cc == 0) ... handle EOF ...;
	if (cc < 0) ... handle error ...;
	for (p = iov, n = iovcnt; n > 0; p++, n--) {
		if (iov->iov_len > cc) {
			iov->iov_base += cc;
			iov->iov_len -= cc;
			break;
		}
		cc -= iov->iov_len;
	}
	if (cc == 0) ... all done ...;
	iov = p;
	iovcnt = n;
	/* now there are iovcnt io vectors undone */

Note that this may modify one of the I/O vectors, hence is not
suitable in some cases.  To be fully general this would have to
copy the vectors in those cases, and work from the copies; then
it would have to free the copies when the whole transfer is done.

This is one reason I elected to ignore readv and writev in my
4BSD stdio rewrite.
-- 
In-Real-Life: Chris Torek, Lawrence Berkeley Lab CSE/EE (+1 415 486 5427)
Berkeley, CA		Domain:	torek@ee.lbl.gov