|Perl Monk, Perl Meditation|
Re^2: powerpc double-double arithmeticby syphilis (Chancellor)
|on Oct 10, 2013 at 02:55 UTC||Need Help??|
In many instances the 16 byte long double provides the same (64-bit) precision as the 12 byte variant ... and the same exponent range, too, I think.
But it's the double-double arrangement that enables contrived comparisons (such as the one I provided) to work.
It would take a data type providing 663 bits of precision to detect that 1.0 + 5e-200 > 1.0, whereas the double-double arrangement of the PowerPC long double stores 1.0 in one double and 5e-200 in the other. The processor then knows that the sum of both is greater than either of the 2 doubles because both doubles are positive and non-zero.
At least, that's how I surmise.
The processor cannot detect that 1.1 + 5e-200 > 1.1 because the "1.1" value gets written over both doubles - and adding 5e-200 then makes no difference to the 2 stored double values.
I was hoping that a printf "%.200Le", ... of the 1.0 + 5e-200 value might print out the actual value of 1.00000...005 but, alas, doesn't even get close (in perl or C).
BTW, the "bitch" part is that, with certain values, you can get rounding errors if you try to print out more than 15 decimal digits of precision (in perl or C).
All in all, an intriguing arrangement - and one that almost works beautifully.