Synchronize with sys/i386/isa/clock.c revision 1.75.
This commit is contained in:
kato
parent
14a9d059f4
commit
53be18e8b1
@ -456,7 +456,7 @@ getit(void)
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void
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DELAY(int n)
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{
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int delta, prev_tick, tick, ticks_left, sec, usec;
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int delta, prev_tick, tick, ticks_left;
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#ifdef DELAYDEBUG
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int getit_calls = 1;
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@ -487,19 +487,30 @@ DELAY(int n)
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* multiplications and divisions to scale the count take a while).
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*/
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prev_tick = getit();
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n -= 20;
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n -= 0; /* XXX actually guess no initial overhead */
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/*
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* Calculate (n * (timer_freq / 1e6)) without using floating point
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* and without any avoidable overflows.
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*/
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sec = n / 1000000;
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usec = n - sec * 1000000;
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ticks_left = sec * timer_freq
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+ usec * (timer_freq / 1000000)
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+ usec * ((timer_freq % 1000000) / 1000) / 1000
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+ usec * (timer_freq % 1000) / 1000000;
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if (n < 0)
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ticks_left = 0; /* XXX timer_freq is unsigned */
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if (n <= 0)
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ticks_left = 0;
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else if (n < 256)
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/*
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* Use fixed point to avoid a slow division by 1000000.
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* 39099 = 1193182 * 2^15 / 10^6 rounded to nearest.
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* 2^15 is the first power of 2 that gives exact results
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* for n between 0 and 256.
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*/
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ticks_left = ((u_int)n * 39099 + (1 << 15) - 1) >> 15;
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else
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/*
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* Don't bother using fixed point, although gcc-2.7.2
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* generates particularly poor code for the long long
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* division, since even the slow way will complete long
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* before the delay is up (unless we're interrupted).
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*/
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ticks_left = ((u_int)n * (long long)timer_freq + 999999)
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/ 1000000;
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while (ticks_left > 0) {
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tick = getit();
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@ -456,7 +456,7 @@ getit(void)
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void
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DELAY(int n)
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{
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int delta, prev_tick, tick, ticks_left, sec, usec;
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int delta, prev_tick, tick, ticks_left;
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#ifdef DELAYDEBUG
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int getit_calls = 1;
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@ -487,19 +487,30 @@ DELAY(int n)
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* multiplications and divisions to scale the count take a while).
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*/
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prev_tick = getit();
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n -= 20;
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n -= 0; /* XXX actually guess no initial overhead */
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/*
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* Calculate (n * (timer_freq / 1e6)) without using floating point
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* and without any avoidable overflows.
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*/
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sec = n / 1000000;
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usec = n - sec * 1000000;
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ticks_left = sec * timer_freq
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+ usec * (timer_freq / 1000000)
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+ usec * ((timer_freq % 1000000) / 1000) / 1000
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+ usec * (timer_freq % 1000) / 1000000;
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if (n < 0)
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ticks_left = 0; /* XXX timer_freq is unsigned */
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if (n <= 0)
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ticks_left = 0;
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else if (n < 256)
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/*
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* Use fixed point to avoid a slow division by 1000000.
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* 39099 = 1193182 * 2^15 / 10^6 rounded to nearest.
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* 2^15 is the first power of 2 that gives exact results
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* for n between 0 and 256.
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*/
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ticks_left = ((u_int)n * 39099 + (1 << 15) - 1) >> 15;
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else
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/*
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* Don't bother using fixed point, although gcc-2.7.2
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* generates particularly poor code for the long long
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* division, since even the slow way will complete long
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* before the delay is up (unless we're interrupted).
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*/
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ticks_left = ((u_int)n * (long long)timer_freq + 999999)
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/ 1000000;
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while (ticks_left > 0) {
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tick = getit();
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@ -456,7 +456,7 @@ getit(void)
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void
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DELAY(int n)
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{
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int delta, prev_tick, tick, ticks_left, sec, usec;
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int delta, prev_tick, tick, ticks_left;
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#ifdef DELAYDEBUG
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int getit_calls = 1;
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@ -487,19 +487,30 @@ DELAY(int n)
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* multiplications and divisions to scale the count take a while).
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*/
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prev_tick = getit();
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n -= 20;
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n -= 0; /* XXX actually guess no initial overhead */
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/*
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* Calculate (n * (timer_freq / 1e6)) without using floating point
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* and without any avoidable overflows.
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*/
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sec = n / 1000000;
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usec = n - sec * 1000000;
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ticks_left = sec * timer_freq
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+ usec * (timer_freq / 1000000)
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+ usec * ((timer_freq % 1000000) / 1000) / 1000
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+ usec * (timer_freq % 1000) / 1000000;
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if (n < 0)
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ticks_left = 0; /* XXX timer_freq is unsigned */
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if (n <= 0)
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ticks_left = 0;
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else if (n < 256)
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/*
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* Use fixed point to avoid a slow division by 1000000.
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* 39099 = 1193182 * 2^15 / 10^6 rounded to nearest.
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* 2^15 is the first power of 2 that gives exact results
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* for n between 0 and 256.
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*/
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ticks_left = ((u_int)n * 39099 + (1 << 15) - 1) >> 15;
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else
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/*
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* Don't bother using fixed point, although gcc-2.7.2
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* generates particularly poor code for the long long
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* division, since even the slow way will complete long
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* before the delay is up (unless we're interrupted).
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*/
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ticks_left = ((u_int)n * (long long)timer_freq + 999999)
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/ 1000000;
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while (ticks_left > 0) {
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tick = getit();
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