diff --git a/sys/pc98/cbus/clock.c b/sys/pc98/cbus/clock.c index 7f2af5a8956a..73453384e2fd 100644 --- a/sys/pc98/cbus/clock.c +++ b/sys/pc98/cbus/clock.c @@ -456,7 +456,7 @@ getit(void) void DELAY(int n) { - int delta, prev_tick, tick, ticks_left, sec, usec; + int delta, prev_tick, tick, ticks_left; #ifdef DELAYDEBUG int getit_calls = 1; @@ -487,19 +487,30 @@ DELAY(int n) * multiplications and divisions to scale the count take a while). */ prev_tick = getit(); - n -= 20; + n -= 0; /* XXX actually guess no initial overhead */ /* * Calculate (n * (timer_freq / 1e6)) without using floating point * and without any avoidable overflows. */ - sec = n / 1000000; - usec = n - sec * 1000000; - ticks_left = sec * timer_freq - + usec * (timer_freq / 1000000) - + usec * ((timer_freq % 1000000) / 1000) / 1000 - + usec * (timer_freq % 1000) / 1000000; - if (n < 0) - ticks_left = 0; /* XXX timer_freq is unsigned */ + 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 + /* + * Don't bother using fixed point, although gcc-2.7.2 + * generates particularly poor code for the long long + * division, since even the slow way will complete long + * before the delay is up (unless we're interrupted). + */ + ticks_left = ((u_int)n * (long long)timer_freq + 999999) + / 1000000; while (ticks_left > 0) { tick = getit(); diff --git a/sys/pc98/cbus/pcrtc.c b/sys/pc98/cbus/pcrtc.c index 7f2af5a8956a..73453384e2fd 100644 --- a/sys/pc98/cbus/pcrtc.c +++ b/sys/pc98/cbus/pcrtc.c @@ -456,7 +456,7 @@ getit(void) void DELAY(int n) { - int delta, prev_tick, tick, ticks_left, sec, usec; + int delta, prev_tick, tick, ticks_left; #ifdef DELAYDEBUG int getit_calls = 1; @@ -487,19 +487,30 @@ DELAY(int n) * multiplications and divisions to scale the count take a while). */ prev_tick = getit(); - n -= 20; + n -= 0; /* XXX actually guess no initial overhead */ /* * Calculate (n * (timer_freq / 1e6)) without using floating point * and without any avoidable overflows. */ - sec = n / 1000000; - usec = n - sec * 1000000; - ticks_left = sec * timer_freq - + usec * (timer_freq / 1000000) - + usec * ((timer_freq % 1000000) / 1000) / 1000 - + usec * (timer_freq % 1000) / 1000000; - if (n < 0) - ticks_left = 0; /* XXX timer_freq is unsigned */ + 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 + /* + * Don't bother using fixed point, although gcc-2.7.2 + * generates particularly poor code for the long long + * division, since even the slow way will complete long + * before the delay is up (unless we're interrupted). + */ + ticks_left = ((u_int)n * (long long)timer_freq + 999999) + / 1000000; while (ticks_left > 0) { tick = getit(); diff --git a/sys/pc98/pc98/clock.c b/sys/pc98/pc98/clock.c index 7f2af5a8956a..73453384e2fd 100644 --- a/sys/pc98/pc98/clock.c +++ b/sys/pc98/pc98/clock.c @@ -456,7 +456,7 @@ getit(void) void DELAY(int n) { - int delta, prev_tick, tick, ticks_left, sec, usec; + int delta, prev_tick, tick, ticks_left; #ifdef DELAYDEBUG int getit_calls = 1; @@ -487,19 +487,30 @@ DELAY(int n) * multiplications and divisions to scale the count take a while). */ prev_tick = getit(); - n -= 20; + n -= 0; /* XXX actually guess no initial overhead */ /* * Calculate (n * (timer_freq / 1e6)) without using floating point * and without any avoidable overflows. */ - sec = n / 1000000; - usec = n - sec * 1000000; - ticks_left = sec * timer_freq - + usec * (timer_freq / 1000000) - + usec * ((timer_freq % 1000000) / 1000) / 1000 - + usec * (timer_freq % 1000) / 1000000; - if (n < 0) - ticks_left = 0; /* XXX timer_freq is unsigned */ + 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 + /* + * Don't bother using fixed point, although gcc-2.7.2 + * generates particularly poor code for the long long + * division, since even the slow way will complete long + * before the delay is up (unless we're interrupted). + */ + ticks_left = ((u_int)n * (long long)timer_freq + 999999) + / 1000000; while (ticks_left > 0) { tick = getit();