Find max integer size that a floating point type can handle without loss of precision

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Published on 2010-04-06T00:52:27Z Indexed on 2010/04/06 1:03 UTC
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Double has range more than a 64-bit integer, but its precision is less dues to its representation (since double is 64-bit as well, it can't fit more actual values). So, when representing larger integers, you start to lose precision in the integer part.

#include <boost/cstdint.hpp>
#include <limits>

template<typename T, typename TFloat>
void
maxint_to_double()
{
    T i = std::numeric_limits<T>::max();
    TFloat d = i;
    std::cout
        << std::fixed
        << i << std::endl
        << d << std::endl;
}

int
main()
{
    maxint_to_double<int, double>();
    maxint_to_double<boost::intmax_t, double>();
    maxint_to_double<int, float>();
    return 0;
}

This prints:

2147483647
2147483647.000000
9223372036854775807
9223372036854775800.000000
2147483647
2147483648.000000

Note how max int can fit into a double without loss of precision and boost::intmax_t (64-bit in this case) cannot. float can't even hold an int.

Now, the question: is there a way in C++ to check if the entire range of a given integer type can fit into a loating point type without loss of precision?

Preferably,

  • it would be a compile-time check that can be used in a static assertion,
  • and would not involve enumerating the constants the compiler should know or can compute.

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