would you mind stating the mathematically correct result (say, to six digits after decimal point)?

According to Math::Decimal128:
C:\>perl -MMath::Decimal128=":all" -le "$n1=Math::Decimal128->new('10' +, 0);$n2=Math::Decimal128->new('1000001', -6);for(1..60000000){$n1 *= + $n2}print $n1;" 114197313013072744502959647597176e-5
Or, to 6 decimal places: 1141973130130727445029596475.971760 which is the same as flexvault got.

With 80-bit long doubles, the calculation came out as:
C:\>perl -MMath::LongDouble=":all" -le "$n1=Math::LongDouble->new('10' +);$n2=Math::LongDouble->new('1.000001');for(1..60000000){$n1 *= $n2}p +rint $n1;" 1.14197313013239314e+027
Or, printing to 28 decimal digits (which implies a level of precision that we don't have):
C:\>perl -MMath::LongDouble=":all" -le "$n1=Math::LongDouble->new('10' +);$n2=Math::LongDouble->new('1.000001');for(1..60000000){$n1 *= $n2}p +rint LDtoSTRP($n1, 28);" 1.141973130132393144424071168e+027
This figure is equivalent to the first one you quoted.

With __float128:
C:\>perl -MMath::Float128=":all" -le "$n1=Math::Float128->new('10');$n +2=Math::Float128->new('1.000001');for(1..60000000){$n1 *= $n2}print $ +n1;" 1.14197313013072744502959647306684e+27
This differs from your second figure - and I don't think there's any valid reason that they should differ.
I'm inclined to think that mine is correct (surprise :-) because it matches up pretty well with the _Decimal128 calculation (where the respective precisions are reasonably close to each other).
Might you have assigned a long double or double value that got cast to a __float128, instead of assigning an actual __float128 value ?

As LanX points out, rounding is involved. The above one-liner programs all employ rounding to nearest with ties to even - the different results of course arising from the different precisions involved.

Cheers,
Rob

In reply to Re^3: Decimal Floating Point (DFP) and does Perl needs DFP? by syphilis
in thread Decimal Floating Point (DFP) and does Perl needs DFP? by flexvault

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