Question

我正在使用Boost 1.55.0和clang 3.5.0以及gcc 4.8.1。

现在我想计算最多256的阶乘（没有精确损失）：

#include <iostream>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/rational.hpp>

int main(){
    using boost::multiprecision::uint128_t;    
    using boost::rational;
    using std::cout;
    using std::endl;

    typedef unsigned long long unsigned_int;
//    typedef uint128_t unsigned_int;    

    rational<unsigned_int> r((unsigned_int)1,

//((unsigned_int)1)<<127);
~(((unsigned_int)-1)>>1));

    unsigned_int n_I = 1;
    cout << "0!:\t\t" << r << endl;
    cout << "1!:\t\t" << r << endl;
    for(unsigned_int i=2; i<257; ++i){
        r *= i;
        cout << i << "!:\t\t" << r << endl;
    }

    return 0;
}

旁注：大型因子在二进制表示中有许多尾随零，因此我从一个值为1 /（2 ^ 127）的有理变量开始。这会自动使分子尽可能小。

我的问题：它不适用于boost multiprecision的uint128_t！但它确实适用于unsigned long long！

这是我的终端输出：

~/ccpp_projects/facultiy $ clang++ -I /usr/local/include/boost-1_55 faculty.cpp -o faculty
In file included from faculty.cpp:51:
In file included from /usr/local/include/boost-1_55/boost/multiprecision/cpp_int.hpp:12:
In file included from /usr/local/include/boost-1_55/boost/multiprecision/number.hpp:22:
In file included from /usr/local/include/boost-1_55/boost/multiprecision/detail/generic_interconvert.hpp:9:
In file included from /usr/local/include/boost-1_55/boost/multiprecision/detail/default_ops.hpp:2073:
/usr/local/include/boost-1_55/boost/multiprecision/detail/no_et_ops.hpp:25:4: error: implicit instantiation of undefined template
      'boost::STATIC_ASSERTION_FAILURE<false>'
   BOOST_STATIC_ASSERT_MSG(is_signed_number<B>::value, "Negating an unsigned type results in ill-defined behavior.");
   ^
/usr/local/include/boost-1_55/boost/static_assert.hpp:36:48: note: expanded from macro 'BOOST_STATIC_ASSERT_MSG'
#     define BOOST_STATIC_ASSERT_MSG( B, Msg ) BOOST_STATIC_ASSERT( B )
                                               ^
/usr/local/include/boost-1_55/boost/static_assert.hpp:169:13: note: expanded from macro 'BOOST_STATIC_ASSERT'
            sizeof(::boost::STATIC_ASSERTION_FAILURE< BOOST_STATIC_ASSERT_BOOL_CAST( __VA_ARGS__ ) >)>\
            ^
/usr/local/include/boost-1_55/boost/rational.hpp:533:15: note: in instantiation of function template specialization
      'boost::multiprecision::operator-<boost::multiprecision::backends::cpp_int_backend<128, 128, 0, 0, void> >' requested here
        num = -num;
              ^
/usr/local/include/boost-1_55/boost/rational.hpp:139:61: note: in instantiation of member function
      'boost::rational<boost::multiprecision::number<boost::multiprecision::backends::cpp_int_backend<128, 128, 0, 0, void>, 0> >::normalize' requested here
    rational(param_type n, param_type d) : num(n), den(d) { normalize(); }
                                                            ^
faculty.cpp:63:28: note: in instantiation of member function 'boost::rational<boost::multiprecision::number<boost::multiprecision::backends::cpp_int_backend<128, 128,
      0, 0, void>, 0> >::rational' requested here
    rational<unsigned_int> r((unsigned_int)1, ~(((unsigned_int)-1)>>1));
                           ^
/usr/local/include/boost-1_55/boost/static_assert.hpp:87:26: note: template is declared here
template <bool x> struct STATIC_ASSERTION_FAILURE;
                         ^
1 error generated.

附录

我刚用g ++编译我的代码，它在那里工作！有没有办法为clang ++禁用BOOST STATIC ASSERT？

Answer 1

normalize()的实现假设翻转基础整数类型的符号（i = -i）是定义的操作。

unsigned long long就是这种情况，uint128_t的不是。

Yould

使用cpp_rational（见 Live On Coliru ）

手动分解2的权力： Live On Coliru ，输出：

0!:     1
1!:     1
2!:     1 x 2^1
3!:     3 x 2^1
4!:     3 x 2^3
...
255!:   62542083004847430224885350954338565259 x 2^247
256!:   62542083004847430224885350954338565259 x 2^255

这可能是你想要的第一个？它会更高效，也可以防止溢出128位。

#include <iostream>
#include <boost/multiprecision/cpp_int.hpp>

int main(){
    using boost::multiprecision::uint128_t;

    uint128_t mantissa = 1;
    unsigned int binary_exponent = 0;

    std::cout << "0!:\t\t" << mantissa << std::endl;
    std::cout << "1!:\t\t" << mantissa << std::endl;

    for(unsigned i=2; i<257; ++i){
        unsigned tmp = i;
        while (tmp && ((tmp % 2) == 0))
        {
            binary_exponent += 1;
            tmp             /= 2;
        }
        mantissa *= tmp;
        std::cout << i << "!:\t\t" << mantissa << " x 2^" << binary_exponent << std::endl;
    }
}

为什么Boost multiprecision不能与Boost有理数库一起使用？

1 个答案: