Hi, The arguments in reverse order (shown below) worked for my Sun system. struct encoding { unsigned long sign:1; unsigned long exp:11; unsigned long mantissa1:20; unsigned long mantissa2; }; Thanks, Chris On 12/21/05, Peder Holt <peder.holt@gmail.com> wrote:
On 12/19/05, Chris Weed <chrisweed@gmail.com> wrote:
Hi, I am trying to simply print a compile-time double using gcc 4.0.0 on Sun. The following prints out : 1.9036e+185
#include <boost/metamath/double.hpp> #include <boost/metamath/double_macros.hpp> #include <iostream>
int main() { typedef BOOST_DOUBLE(3.3) D; D d; std::cout << (double)d << std::endl; return 0; }
Am I doing something wrong? Chris
This has probably got something to do with the way I translate a mpl::double_ to a double. I use a union to represent the mapping between a double and the mantissa,exponent and sign (represented as integers) This mapping is most likely different for different systems, so we need to either device a better scheme for converting from compile-time to runtime double, or use different mappings for different systems.
The relevant code is found in: boost\mpl\math\double_\aux_\get_value.hpp
struct encoding { unsigned long mantissa2; unsigned long mantissa1:20; unsigned long exp:11; unsigned long sign:1; };
union converter_ { double value; encoding encode; };
template <typename T> double get_value(const T& arg) { converter_ converter; converter.encode.sign = T::sign; converter.encode.exp = T::exponent + 1023; converter.encode.mantissa1 = T::mantissa::part1 >> 10; converter.encode.mantissa2 = (T::mantissa::part1 << 22) | (T::mantissa::part2 >> 9); return converter.value; }
Try fiddling with the order of the elements in the encoding struct, e.g. struct encoding { unsigned long mantissa1:20; unsigned long exp:11; unsigned long sign:1; //Move this to the end unsigned long mantissa2; };
On the other hand, this encoding is probably documented somewhere...
Regards, Peder
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