> What's the reasoning behind your assertion ? 

Exponentiation laws.

E.g. x^10 = x^8 * x^2 and x^8 = ((x^2)^2)^2

Just because 10 is 1010 binary, hence < 4*2 multiplications.

6e7 should have a 26 bit representation so <52 multiplications needed.

Of course there are even faster implementations which are less exact. I think that's what you are observing.

edit
Of course a binary calculation should be done with integer x=1000001 without fraction, dividing later thru 10^(6*60) should be easy enough.

Cheers Rolf

PS: Je suis Charlie!

update

proof of concept -> here

In reply to Re^6: Decimal Floating Point (DFP) and does Perl needs DFP? by LanX
in thread Decimal Floating Point (DFP) and does Perl needs DFP? by flexvault

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