On Mar 14, 2011, at 16:22, York Sun wrote: > On Wed, 2011-02-23 at 11:35 -0500, Kyle Moffett wrote: >> + * Now divide by 5^12 and track the 32-bit remainder, then divide >> + * by 2*(2^12) using shifts (and updating the remainder). >> + */ >> + clks_rem = do_div(clks, UL_5pow12); >> + clks_rem <<= 13; > > Shouldn't this be clks_rem >>= 13 ? >> >> + clks_rem |= clks & (UL_2pow13-1); >> + clks >>= 13; >> + >> + /* If we had a remainder, then round up */ >> + if (clks_rem) >> clks++; >> - }
Since I'm dividing a second time, the old remainder value represents the high bits of the new remainder value and therefore needs to be left-shifted, then the now-empty low bits of the remainder are taken from the bits of "clks" which get shifted away. Example: Say I want to divide 1999999999999 (IE: 2*10^^12 - 1) by 2000000000000 (2*10^^12). Obviously the dividend is less than the divisor (by 1), so the result should be 0 and the remainder should be equal to the dividend. So my initial value is: clks = 1999999999999; Now I divide by 5^^12: clks_rem = do_div(clks, 244140625); /* This number is 5pow12 */ The results are: clks_rem == 244140624 clks == 8191 Now I shift left: clks_rem <<= 13; And get: clks_rem == 1999999991808 Finally, I copy the low bits of clks to clks_rem and shift them out of clks: clks_rem |= clks & (UL_2pow13-1); clks >>= 13 The result is as expected: clks_rem == 1999999999999; clks == 0; Cheers, Kyle Moffett _______________________________________________ U-Boot mailing list U-Boot@lists.denx.de http://lists.denx.de/mailman/listinfo/u-boot
