void
DELAY(int n)
{
if (delay_tc(n))
return;
init_ops.early_delay(n);
}
/*
* Wait "n" microseconds.
* Relies on timer 1 counting down from (i8254_freq / hz)
* Note: timer had better have been programmed before this is first used!
*/
void
DELAY(int n)
{
int delta, prev_tick, tick, ticks_left;
#ifdef DELAYDEBUG
int getit_calls = 1;
int n1;
static int state = 0;
if (state == 0) {
state = 1;
for (n1 = 1; n1 <= 10000000; n1 *= 10)
DELAY(n1);
state = 2;
}
if (state == 1)
printf("DELAY(%d)...", n);
#else
if (delay_tc(n))
return;
#endif
/*
* Read the counter first, so that the rest of the setup overhead is
* counted. Guess the initial overhead is 20 usec (on most systems it
* takes about 1.5 usec for each of the i/o's in getit(). The loop
* takes about 6 usec on a 486/33 and 13 usec on a 386/20. The
* multiplications and divisions to scale the count take a while).
*
* However, if ddb is active then use a fake counter since reading
* the i8254 counter involves acquiring a lock. ddb must not do
* locking for many reasons, but it calls here for at least atkbd
* input.
*/
#ifdef KDB
if (kdb_active)
prev_tick = 1;
else
#endif
prev_tick = getit();
n -= 0; /* XXX actually guess no initial overhead */
/*
* Calculate (n * (i8254_freq / 1e6)) without using floating point
* and without any avoidable overflows.
*/
if (n <= 0)
ticks_left = 0;
else if (n < 256)
/*
* Use fixed point to avoid a slow division by 1000000.
* 39099 = 1193182 * 2^15 / 10^6 rounded to nearest.
* 2^15 is the first power of 2 that gives exact results
* for n between 0 and 256.
*/
ticks_left = ((u_int)n * 39099 + (1 << 15) - 1) >> 15;
else
void
DELAY(int n)
{
TSENTER();
if (delay_tc(n)) {
TSEXIT();
return;
}
init_ops.early_delay(n);
TSEXIT();
}