boost rational is tripping up one of my examples in safe numerics:
// solution: use safe integer in rational definition
using safe_rational = boost::rational<
boost::safe_numerics::safe<int>
>;
// use rationals created with safe_t
const safe_rational sc {1, std::numeric_limits<int>::max()};
std::cout << "c = " << sc << std::endl;
const safe_rational sd {1, 2};
std::cout << "d = " << sd << std::endl;
std::cout << "c * d = ";
try {
// multiply them. This will overflow
std::cout << sc * sd << std::endl; // use of overload
operator * is ambiguous.
rational.hpp contains - among other things, the following definitions
for the * operator:
template
BOOST_CXX14_CONSTEXPR
inline typename boost::enable_if_c <
rational_detail::is_compatible_integer::value ||
is_same::value, rational<IntType> >::type
operator * (const rational<IntType>& a, const Arg& b)
{
rational<IntType> t(a);
return t *= b;
}
template
BOOST_CXX14_CONSTEXPR
inline typename boost::enable_if_c <
rational_detail::is_compatible_integer::value,
rational<IntType> >::type
operator * (const Arg& b, const rational<IntType>& a)
{
rational<IntType> t(a);
return t *= b;
}
Soooooo - it seems that sc * sd will match both of the above
definitions. Its unclear what the purpose of these two different
overloads are. They look pretty similar to me.
the definition for is_compatible_integer.
namespace rational_detail{
template
struct is_compatible_integer;
template
struct is_compatible_integer::type>
{
BOOST_STATIC_CONSTANT(bool, value =
((std::numeric_limits<FromInt>::is_specialized
&& std::numeric_limits<FromInt>::is_integer
&& (std::numeric_limits<FromInt>::digits
<=std::numeric_limits<ToInt>::digits)
&& (std::numeric_limits<FromInt>::radix ==
std::numeric_limits<ToInt>::radix)
&& ((std::numeric_limits<FromInt>::is_signed == false) ||
(std::numeric_limits<ToInt>::is_signed == true))
&& is_convertible::value)
|| is_same::value)
|| (is_class<ToInt>::value && is_class<FromInt>::value &&
is_convertible::value));
};
...
In fact, since it looks like if FromInt and ToInt are the same types (as
they are in my case, one will always got more than one match. Is anyone
(John Maddock - where are you?) able to shed some light on this for me?
Robert Ramey
