CODEWITHSUNDEEP
c++boostfloating-pointprecisionmultiprecision
maxschlepzig
maxschlepzig

Reputation: 39027

What is the precision of cpp_dec_float_50?

Looking at the name and the Boost Multiprecision documentation I would expect that the cpp_dec_float_50 datatype has a precision of 50 decimal digits:

Using typedef cpp_dec_float_50 hides the complexity of multiprecision to allow us to define variables with 50 decimal digit precision just like built-in double.

(Although I don't understand the comparison with double - I mean double usually implements binary floating point arithmetic, not decimal floating point arithmetic.)

This is also matched by the output of following code (except for the double part, but this is expected):

cout << std::numeric_limits<boost::multiprecision::cpp_dec_float_50>::digits10
  << '\n';
// -> 50
cout << std::numeric_limits<double>::digits10 << '\n';
// -> 15

But why does following code print 74 digits then?

#include <boost/multiprecision/cpp_dec_float.hpp>

// "12" repeated 50 times, decimal point after the 10th digit
boost::multiprecision::cpp_dec_float_50 d("1212121212.121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212121212");
cout << d.convert_to<string>() << '\n';
// Expected output: 50 digits
// Actual output: 74 digits
// -> 1212121212.1212121212121212121212121212121212121212121212121212121212121212

The str() member function works as expected, e.g.

cout << d.str(50) << '\n';

does only print 50 digits - where it is documented as:

Returns the number formatted as a string, with at least precision digits, and in scientific format if scientific is true.

Upvotes: 3

Views: 1681

Answers (1)

sehe
sehe

Reputation: 392833

What you are seeing is likely related to the guard digits used internally. The reason is that even decimal representation has limited accuracy (think ("100.0" / "3.0") * "3.0").

In order to get reasonable rounding errors during calculations, the stored precision will be more than the guaranteed precision.

In summary: always be specific about your expected precision. In your example d.str(50) would do.

(In realistic scenarios, you should want to track the precision of your inputs and deduce the precision on your outputs. Most often, people just reserve surplus precision and only print the part they're interested in)

Upvotes: 3

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