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From: Thorsten O. <ad...@th...> - 2018-05-18 14:58:32
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On Freitag, 18. Mai 2018 16:33:28 CEST David Gálvez wrote: > the loop count would be 0 > for any given value of nanoseconds below 1000, so the loop count will > be round up to 1 to get at least the delay requested. Yes, but if the loop count is so low that it is not really precise for microsecs, whats the purpose then computing it based on nanosecs? Even a 060@95Mhz is not 1000 times faster than a 68000. Also, given that one of the factors is already essentially shifted right by 28 bits, that leaves only 4 bits, multiplied by the ns value, then divided by 1000, and thus only a range of 1-16 loops, if i read the code correctly. |