Approximate the divide by multiplying by 3433/4096. This is faster than
the previous code, and more accurate. Also avoid saving and restoring registers. (The multiplication algorithm was generated by GCC.)
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915336b97c
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@ -1,4 +1,4 @@
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/* $NetBSD: microtime.s,v 1.12 1994/11/05 02:16:26 mycroft Exp $ */
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/* $NetBSD: microtime.s,v 1.13 1994/11/06 20:33:35 mycroft Exp $ */
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/*-
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* Copyright (c) 1993 The Regents of the University of California.
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@ -46,26 +46,17 @@
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*/
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#ifndef HZ
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ENTRY(microtime)
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pushl %edi
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pushl %esi
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pushl %ebx
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movl $_time,%ebx # get pointer to time
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cli # disable interrupts
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movl (%ebx),%edi # sec = time.tv_sec
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movl 4(%ebx),%esi # usec = time.tv_usec
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movb $(TIMER_SEL0|TIMER_LATCH),%al
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outb %al,$TIMER_MODE # latch timer 0's counter
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# Read counter value into ebx, LSB first
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xorl %ebx,%ebx
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# Read counter value into ecx, LSB first
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xorl %ecx,%ecx
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inb $TIMER_CNTR0,%al
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movb %al,%bl
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movb %al,%cl
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inb $TIMER_CNTR0,%al
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movb %al,%bh
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movb %al,%ch
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# Now check for counter overflow. This is tricky because the
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# timer chip doesn't let us atomically read the current counter
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@ -79,61 +70,60 @@ ENTRY(microtime)
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# from the IRR, and mistakenly add a correction to the "close
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# to zero" value.
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#
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# We compare the counter value to heuristic constant 12.
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# We compare the counter value to the heuristic constant 12.
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# If the counter value is less than this, we assume the counter
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# didn't overflow between disabling interrupts above and latching
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# the counter value. For example, we assume that the above 10 or so
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# didn't overflow between disabling clock interrupts and latching
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# the counter value above. For example, we assume that the first 3
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# instructions take less than 12 microseconds to execute.
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#
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# We used to check for overflow only if the value read was close to
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# (We used to check for overflow only if the value read was close to
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# the timer limit, but this doesn't work very well if we're at the
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# clock's ipl or higher.
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# clock's ipl or higher.)
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#
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# Otherwise, the counter might have overflowed. We check for this
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# condition by reading the interrupt request register out of the ICU.
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# If it overflowed, we add in one clock period.
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movl $11932,%edx # timer limit
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movl $11932,%edx # counter limit
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testb $IRQ_BIT(0),_ipending + IRQ_BYTE(0)
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jnz 1f
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cmpl $12,%ebx # check for potential overflow
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cmpl $12,%ecx # check for potential overflow
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jbe 2f
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inb $IO_ICU1,%al # read IRR in ICU
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testb $IRQ_BIT(0),%al # is a timer interrupt pending?
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jz 2f
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1: subl %edx,%ebx # add another tick
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1: subl %edx,%ecx # add another tick
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2: subl %ebx,%edx # subtract counter value from limit
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2: subl %ecx,%edx # subtract counter value from counter limit
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# Divide by 1193280/1000000. We use a fast approximation of 4096/3433.
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# For values of hz more than 100, this has a maximum error of 2us.
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leal (%edx,%edx,2),%eax # a = 3d
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leal (%edx,%eax,4),%eax # a = 4a + d = 13d
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movl %eax,%ecx
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shll $5,%ecx
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addl %ecx,%eax # a = 33a = 429d
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leal (%edx,%eax,8),%eax # a = 8a + d = 3433d
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shrl $12,%eax # a = a/4096 = 3433d/4096
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movl time,%edx # get time.tv_sec
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addl time+4,%eax # add time.tv_usec
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sti # enable interrupts
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movl %edx,%eax # movl %edx,%eax; imull $1000,%eax,%eax
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sall $10,%eax
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sall $3,%edx
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subl %edx,%eax
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sall $1,%edx
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subl %edx,%eax
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xorl %edx,%edx # zero extend eax for div
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movl $1193,%ecx
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idivl %ecx # convert to usecs: mult by 1000/1193
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addl %eax,%esi # add counter usecs to time.tv_usec
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cmpl $1000000,%esi # carry in timeval?
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jl 3f
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subl $1000000,%esi # adjust usec
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incl %edi # bump sec
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cmpl $1000000,%eax # carry in timeval?
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jb 3f
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subl $1000000,%eax # adjust usec
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incl %edx # bump sec
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3: movl 16(%esp),%ecx # load timeval pointer arg
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movl %edi,(%ecx) # tvp->tv_sec = sec
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movl %esi,4(%ecx) # tvp->tv_usec = usec
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movl %edx,(%ecx) # tvp->tv_sec = sec
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movl %eax,4(%ecx) # tvp->tv_usec = usec
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popl %ebx # restore regs
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popl %esi
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popl %edi
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ret
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#endif
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