为什么Boost多precision与升压有理数库工作吗？ [英] Why does Boost multiprecision not work with Boost rational number library?

问题描述

我使用升压1.55.0铿锵3.5.0和gcc 4.8.1。

I'm using Boost 1.55.0 with clang 3.5.0 and gcc 4.8.1.

现在我想计算阶乘多达256个（没有precisionloss）：

Now I would like to compute factorials up to 256 (with no precisionloss):

#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;
}

边注：大阶乘在二进制重新presentation许多尾随零，所以我开始用1 /（2 ^ 127）的价值理性的变量。此自动保持分子尽可能小

Side note: Large factorials have many trailing zeros in the binary representation, so I start with a rational variable with the value of 1/(2^127). This automatically keeps the numerator as small as possible.

我的问题：
它不以 uint128_t 从boost多precision工作！
但它确实与工作无符号长长！

My problem: It does not work with uint128_t from boost multiprecision! but it does work with unsigned long long!

下面是我的终端输出：

~/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

我刚刚编译我的code与G ++和它的作品在那里！
是否有某种方式来禁用BOOST静态断言为铛++？

I just compiled my code with g++ and it works there! Is there some way to disable BOOST STATIC ASSERT for clang++?

推荐答案

的实施正常化（）假设翻转号（ I = -i ）的底层整数类型的是定义的操作。

The implementation of normalize() assumed that flipping the sign (i = -i) of the underlying integer type is a defined operation.

这是的情况下无符号长长，但是的不的为 uint128_t 。

Yould

1. 使用 cpp_rational （看到它的的住在Coliru ）

1. use cpp_rational (see it Live On Coliru)

手工分解出2的幂： 住在Coliru ，输出：

manually factor out the powers of 2: Live On Coliru, output:

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

这是有可能你首先想要的是什么？这将是更有效，也prevent溢满了128位。

This is likely what you wanted in the first place? It will be more efficient and also prevent overflowing the 128 bit.

#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;
    }
}

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