具有两个cpp_int值的boost :: multiprecision :: pow

Question

Is there a way to use boost::multiprecision::pow with two cpp_int values as the parameters? All I can find in the documentation has the parameters of type cpp_int for the base and int for the exponent. This seems very limiting.

Answer 1

Comment: How is that limiting? Will you be raising to powers above MAXINT? How?

Q :
What do you mean 'how'? Isn't the point of multiprecision libraries to handle enormous numbers like this?

No. Enormous numbers like this rarely make any sense at lossless precisions.

Lets say we start out with a reasonably small number, like 10 . The smallest exponent that doesn't fit in a 64-bit integer would be 2^64. So, the number is 10^(2^64), which is roughly 18446744073709551617 decimal digits ≈ 1.84467×10^19 decimal digits.

To print that you'd need paper weighing roughly 1.4757×10^11 metric tons, assuming 5,000 digits per 80g page . That's roughly the equivalent of total biomass on Earth (≈ 8×10^13 kg).

Now, of course you're not silly, and you won't print it! You only need to fit it in RAM, which is why you have started crowd funding for your 7.6598 exabytes of RAM. Not to mention the power supply for it, because powering that for one hour will cost take around 7 gigawatt hours, which is comparable to half the energy yield of Little Boy nuclear bomb.

What Can You Do

Boost Multiprecision does allow you to use exact and lossless representation of huge integral number, but the resources of your systems limit the total capacity.

As shown, exponents exceeding 64 bit integers are not sensible for these types.

You can, of course, use a decimal/binary floating point representation at arbitrary precision (still honouring the constraints of physics and economics, of course), like boost::multiprecision::mpf_float_1000 .

Answer 2

您可以使用boost :: multiprecision :: float和相应的boost :: multi precision :: pow实现。