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Implement a simple `fairness' algorithm which doesn't cost too much in 

performance.
This commit is contained in:
mycroft 1994-04-20 07:55:38 +00:00
parent 893be926ec
commit cf78b04c1a
1 changed files with 8 additions and 10 deletions
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18
sys/arch/i386/isa/fd.c
View File

@ -35,7 +35,7 @@
* SUCH DAMAGE.
*
* from: @(#)fd.c 7.4 (Berkeley) 5/25/91
* $Id: fd.c,v 1.40 1994/04/20 07:23:52 mycroft Exp $
* $Id: fd.c,v 1.41 1994/04/20 07:55:38 mycroft Exp $
*/
#include <sys/param.h>
@ -169,6 +169,7 @@ struct fd_softc {
int sc_skip; /* bytes transferred so far */
int sc_track; /* where we think the head is */
int sc_nblks; /* number of blocks tranferring */
int sc_ops; /* I/O operations completed */
daddr_t sc_blkno; /* starting block number */
};
@ -773,6 +774,7 @@ again:
bp = fd->sc_q.b_actf;
if (!bp) {
/* nothing queued on this drive; try next */
fd->sc_ops = 0;
TAILQ_REMOVE(&fdc->sc_drives, fd, sc_drivechain);
fd->sc_q.b_active = 0;
goto again;
@ -1057,24 +1059,20 @@ fdfinish(fd, bp)
{
struct fdc_softc *fdc = (void *)fd->sc_dev.dv_parent;
#if 0
/*
* This might seem like a good idea, but each drive switch takes .25
* second because we can't keep both motors running at the same time
* and we have to wait for the new one to stabilize for each I/O.
*/
/*
* Move this drive to the end of the queue to give others a `fair'
* chance.
* chance. We only force a switch if N operations are completed while
* another drive is waiting to be serviced, since there is a long motor
* startup delay whenever we switch.
*/
if (fd->sc_drivechain.tqe_next) {
if (fd->sc_drivechain.tqe_next && ++fd->sc_ops >= 8) {
fd->sc_ops = 0;
TAILQ_REMOVE(&fdc->sc_drives, fd, sc_drivechain);
if (bp->b_actf) {
TAILQ_INSERT_TAIL(&fdc->sc_drives, fd, sc_drivechain);
} else
fd->sc_q.b_active = 0;
}
#endif
bp->b_resid = bp->b_bcount - fd->sc_skip;
fd->sc_skip = 0;
fd->sc_q.b_actf = bp->b_actf;