[Boost-bugs] [Boost C++ Libraries] #10085: Improve cpp_rational::convert_to<double>

Subject: [Boost-bugs] [Boost C++ Libraries] #10085: Improve cpp_rational::convert_to<double>
From: Boost C++ Libraries (noreply_at_[hidden])
Date: 2014-06-01 20:29:41

#10085: Improve cpp_rational::convert_to<double>
------------------------------+----------------------------
 Reporter:  marc.glisse@…     |      Owner:  johnmaddock
     Type:  Feature Requests  |     Status:  new
Milestone:  To Be Determined  |  Component:  multiprecision
  Version:  Boost 1.55.0      |   Severity:  Problem
 Keywords:                    |
------------------------------+----------------------------
 Hello,

 the following program prints -nan. If I convert to long double, it prints
 1 as expected, but the same problem happens for slightly bigger numerators
 and denominators. Worse, I can make it return inf or 0 for numbers close
 to 10 or 0.1. Instead of converting numerator and denominator to double
 and dividing, it would work better if you had some kind of frexp function
 that you could apply to both, so the exponents could cancel out without
 overflowing (that might actually be the same as converting to a
 cpp_float<53> if it existed).

 By the way, it would be nice to document what rounding we can expect from
 the conversion to double. For cpp_int, can I assume the result is within 1
 ulp, rounded towards 0? .5 ulp, rounded to nearest? My goal is to get an
 interval of double that contains the number
 (convert_to<boost::numeric::interval_lib::interval<double,P>> basically,
 except that I don't use boost's type for intervals).

 {{{#!c++
 #include <iostream>
 #include <limits>
 #include <boost/multiprecision/cpp_int.hpp>
 typedef boost::multiprecision::cpp_rational NT;

 int main(){
   NT x (std::numeric_limits<double>::min());
   x *= x;
   x += 1;
   std::cout << x.convert_to<double>() << '\n';
 }
 }}}

 PS: thank you for the quick resolution to the previous ticket!

-- 
Ticket URL: <https://svn.boost.org/trac/boost/ticket/10085>
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