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(); }