How to raise the precision of pow in C++ for large numbers (10^19)?

Question

I am looking to calculate 9^19. my code is:

    cout.setf(ios::fixed, ios::floatfield);
    cout.setf(ios::showpoint);
    cout<<pow(9,19)<<endl;

The result has the last 2 digits equal to 0: 1350851717672992000. In Python,9**19 got me 1350851717672992089L . Seems a floating point issue. How could I raise the precision for pow? or how to preform a better precision power than pow?

I am compiling with gcc version 4.8.2.

You are using double. Using long double or long long would work better (depending on the platform). — Marc Glisse, May 13 '15 at 12:49
Python has arbitrary precision arithmetic for integers. With C++ if you want that you'll have to use some 3rd party "bignum" library, or create such yourself. — Cheers and hth. - Alf, May 13 '15 at 12:51
@MarcGlisse it worked by casting entries to `( long double)`. — Assem, May 13 '15 at 12:51
Unfortunately MSVC doesn't support 80-bit double but using a 64-bit int will give the correct result (not with `pow()` of course). — Weather Vane, May 13 '15 at 12:54

score 9 · Accepted Answer · answered May 13 '15 at 12:50

It is indeed a floating-point issue: a typical 64-bit double only gives 53 bits, or about 15 decimal digits, of precision.

You might (or might not) get more precision from long double. Or, for integers up to about 10¹⁹, you could use uint64_t. Otherwise, there's no standard type with more precision: you'll need a library such as GMP or Boost.Multiprecision.

score 1 · Answer 2 · answered May 13 '15 at 12:59

1

The precision of double is around 16 decimal digits.

That means, that only the first 16 decimal digits are exact, the rest of the digits are as good as noise.

answered May 13 '15 at 12:59

Maxim Egorushkin

131,725
17
180
271

How to raise the precision of pow in C++ for large numbers (10^19)?

2 Answers2