Messages in this thread | | | Date | Sat, 5 Jun 2004 17:23:26 +0200 | From | Dominik Brodowski <> | Subject | Re: Too much error in __const_udelay() ? |
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Hi,
> However I've started to see some problems w/ 2.6 and USB on x440/x445s, > both of which use the 100Mhz cyclone time source. Further digging has > pointed to the fact that certain important udelay()s in the USB > subsystem aren't actually waiting long enough.
Certain? AFAICS _no_ call to a delay routine actually passed a big enough argument. Or am I missing something? Also, __ndelay seems to be affected as well: it returns zero for 550 nsec even for the TSC variant in your test.c.
> So I'm no math wiz. What's the proper fix here?
Below are three changes I'd like to discuss. I'll build a fresh kernel with all three changes enabled + PM_TIMER soon.
Change 1:
Move the multiplication with HZ up into the mull instruction:
unsigned long __const_udelay(unsigned long xloops) { int d0; __asm__("mull %0" :"=d" (xloops), "=&a" (d0) :"1" (xloops),"0" (LPJ * HZ)); return __delay(xloops); }
1 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 99 1 usec: LPJ: 1500000 __udelay: 1000 vs my_udelay: 1499
2 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 199 2 usec: LPJ: 1500000 __udelay: 2000 vs my_udelay: 2999
5 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 499 5 usec: LPJ: 1500000 __udelay: 7000 vs my_udelay: 7498
10 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 999 10 usec: LPJ: 1500000 __udelay: 14000 vs my_udelay: 14996
20 usec: LPJ: 100000 __udelay: 1000 vs my_udelay: 1999 20 usec: LPJ: 1500000 __udelay: 29000 vs my_udelay: 29993
50 usec: LPJ: 100000 __udelay: 4000 vs my_udelay: 4998 50 usec: LPJ: 1500000 __udelay: 74000 vs my_udelay: 74983
100 usec: LPJ: 100000 __udelay: 9000 vs my_udelay: 9997 100 usec: LPJ: 1500000 __udelay: 149000 vs my_udelay: 149966
20000 usec: LPJ: 100000 __udelay: 1999000 vs my_udelay: 1999549 20000 usec: LPJ: 1500000 __udelay: 29993000 vs my_udelay: 29993243
Change 2:
Round up in __udelay. While it can be argued that some time is also spent in the delay functions, it's better to spend _at least_ the specified time sleeping, in my humble opinion.
- return __const_udelay2(usecs * 0x000010c6); /* 2**32 / 1000000 */ + return __const_udelay2(usecs * 0x000010c7); /* 2**32 / 1000000 (rounded up)*/
1 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 100 1 usec: LPJ: 1500000 __udelay: 1000 vs my_udelay: 1500
2 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 200 2 usec: LPJ: 1500000 __udelay: 2000 vs my_udelay: 3000
5 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 500 5 usec: LPJ: 1500000 __udelay: 7000 vs my_udelay: 7500
10 usec: LPJ: 100000 __udelay: 0 vs my_udelay: 1000 10 usec: LPJ: 1500000 __udelay: 14000 vs my_udelay: 15000
20 usec: LPJ: 100000 __udelay: 1000 vs my_udelay: 2000 20 usec: LPJ: 1500000 __udelay: 29000 vs my_udelay: 30000
50 usec: LPJ: 100000 __udelay: 4000 vs my_udelay: 5000 50 usec: LPJ: 1500000 __udelay: 74000 vs my_udelay: 75000
100 usec: LPJ: 100000 __udelay: 9000 vs my_udelay: 10000 100 usec: LPJ: 1500000 __udelay: 149000 vs my_udelay: 150001
20000 usec: LPJ: 100000 __udelay: 1999000 vs my_udelay: 2000015 20000 usec: LPJ: 1500000 __udelay: 29993000 vs my_udelay: 30000228
Change 3:
Asserting at least 1 loop is spent: in really small ndelay() calls to low-mhz timers, this might be better.
return __delay(xloops ? xloops : 1);
Before: 1 nsec: LPJ: 100000 __ndelay: 0 vs my_udelay: 0 2 nsec: LPJ: 100000 __ndelay: 0 vs my_udelay: 0 5 nsec: LPJ: 100000 __ndelay: 0 vs my_udelay: 0 10 nsec: LPJ: 100000 __ndelay: 0 vs my_udelay: 1 20 nsec: LPJ: 100000 __udelay: 0 vs my_udelay: 2 50 nsec: LPJ: 100000 __ndelay: 0 vs my_udelay: 5
After: 1 nsec: LPJ: 100000 __udelay: 0 vs my_udelay: 1 2 nsec: LPJ: 100000 __udelay: 0 vs my_udelay: 1 5 nsec: LPJ: 100000 __udelay: 0 vs my_udelay: 1 10 nsec: LPJ: 100000 __udelay: 0 vs my_udelay: 1 20 nsec: LPJ: 100000 __udelay: 0 vs my_udelay: 2 50 nsec: LPJ: 100000 __udelay: 0 vs my_udelay: 5
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