在 c++ 中使用按位运算隔离双精度的小数部分

Question

我被困在现代数字软件开发 class 的任务上。

Function 原型（假设 x = 6.5）：

//returns the IEEE fractional part of x as a decimal floating point number. You must convert binary to decimal.
inline double fraction(double x) {}

我得到了什么：

inline double fraction(double x)
{
    // Get the fraction
    unsigned long long frac_mask = (1u << 52) - 1;                      // Get 52 1's
    unsigned long long xint = *reinterpret_cast<long long*>(&x);        // Interpret x's bits as an int
    unsigned long long frac_num = xint & frac_mask;                                 // Get the fraction as an int
    double fraction = double(frac_num) / double(2u << 52);              // Divide frac_num by 2^52

    return fraction;

    /* This code works, but is not what is specified:
        double fraction = x / pow(2, exponent(x));
        fraction = fmod(fraction, 1);
        return fraction;
    */
}

我不断得到一个NaN。 我正在寻找的答案是 0.625。 我有点绝望地迷失了。 任何帮助深表感谢。

我能够使用以下 function 成功隔离双精度数的指数：

inline int exponent(double x) //returns the unbiased(true) binary exponent of x as a decimal integer. Remember that subnormals are a special case. Consider 0 to be a subnormal.
{
    if (x == 0.0)
        return -1022;
    else if (isnan(x))
        return 1024;

    // Get the exponent
    unsigned long long exp_mask = (1u << 11) - 1;                       // Get eleven 1's
    exp_mask <<= 52;                                                    // Move into place
    unsigned long long xint = *reinterpret_cast<long long*>(&x);        // Interpret x's bits as an int
    unsigned long long exp_bits = xint & exp_mask;                      // Get the exponent bits
    unsigned long long exp = exp_bits >> 52;                            // Get the exponent as a number
    return exp -1023;
}

我很困惑为什么指数逻辑有效，但分数不会。

Answer 1

您正在将unsigned （可能是32位）与需要64位的值混合。

例如， frac_num只有32位，使用long或long long ... [或uint64_t ，这是获得64位值的更可靠方法。

inline double fraction(double x)
{
    // Get the fraction
    uint64_t frac_mask = (1ul << 52) - 1;                      // Get 52 1's
//    uint64_t xint = *reinterpret_cast<uint64_t*>(&x);        // Interpret x's bits as an int
      uint64_t xint; 
      memcpy(&xint, &x, sizeof(xint));        // Interpret x's bits as an int
    int64_t frac_num = xint & frac_mask;                                    // Get the fraction as an int
    frac_num += 1ul << 52; // Add hidden bit.
    double fraction = double(frac_num) / double(2ul << 52);              // Divide frac_num by 2^52

    return fraction;
}

注意向1u和2u添加l ，以确保它们很long ，并且。 您需要包含cstdint才能获得大小整数。

编辑：那当然只是给你一个分数形式的尾数。 小数点可以是位1023和-1023之间的任何位置，这意味着只有-1和+1之间的值才能得到正确的结果。

使用上面代码的完整示例[+ some printouts]

#include <cstdint>
#include <iostream>
#include <cstring>

inline double fraction(double x)
{
    // Get the fraction
    uint64_t frac_mask = (1ul << 52) - 1;                      // Get 52 1's
    std::cout << "mask=" << std::hex << frac_mask << std::endl;
//    uint64_t xint = *reinterpret_cast<uint64_t*>(&x);        // Interpret x's bits as an int
      uint64_t xint; 
      memcpy(&xint, &x, sizeof(xint));        // Interpret x's bits as an int
    int64_t frac_num = xint & frac_mask;                                    // Get the fraction as an int

    frac_num += 1ul << 52; // Add hidden bit.
    std::cout << "xint=" << std::hex << xint << " num=" << std::hex << frac_num << std::endl;
    double fraction = double(frac_num) / double(2ul << 52);              // Divide frac_num by 2^52

    return fraction;
}


int main()
{
    double a = 0.5;
    double b = 0.75;
    double d = 6.5;
    double e = 4.711;


    double fa  = fraction(a);
    double fb  = fraction(b);
    double fd  = fraction(d);
    double fe  = fraction(e);

    std::cout << "fa=" << std::fixed << fa << " fb=" << fb << std::endl;
    std::cout << "fd=" << std::fixed << fd << " fe=" << fe << std::endl;
}

重新运行以上内容：

mask=fffffffffffff
xint=3fe0000000000000 num=10000000000000
mask=fffffffffffff
xint=3fe8000000000000 num=18000000000000
mask=fffffffffffff
xint=401a000000000000 num=1a000000000000
mask=fffffffffffff
xint=4012d810624dd2f2 num=12d810624dd2f2
fa=0.500000 fb=0.750000
fd=0.812500 fe=0.588875

请注意，如果将4.711除以2几次[确切地说是3次]，则得到0.588875，如果将6.5除以8（或得到2除以3），则得到0.8125

我需要去睡觉，但你基本上必须考虑指数来计算浮点数的分数。 或者简单地转换为整数，然后减去它 - 只要它在范围内。

Answer 2

代码：在线试用

// bit_cast, bit_width
#include <bit>
// assert
#include <cassert>
// uint64_t
#include <cstdint>

[[nodiscard]]
constexpr auto Frac(double x)
    noexcept -> double
{
    using Bits = std::uint64_t;
    constexpr Bits s_sign_bit_count{ 1ull };
    constexpr Bits s_exponent_bit_count{ 11ull };
    constexpr Bits s_mantissa_bit_count{ 52ull };
    constexpr Bits s_sign_max{ (1ull << s_sign_bit_count) - 1ull };
    constexpr Bits s_exponent_max{ (1ull << s_exponent_bit_count) - 1ull };
    constexpr Bits s_mantissa_max{ (1ull << s_mantissa_bit_count) - 1ull };
    constexpr Bits s_sign_mask{ s_sign_max << (s_exponent_bit_count + s_mantissa_bit_count) };
    constexpr Bits s_exponent_mask{ s_exponent_max << s_mantissa_bit_count };
    constexpr Bits s_mantissa_mask{ s_mantissa_max };
    constexpr Bits s_exponent_bias{ (1ull << (s_exponent_bit_count - 1ull)) - 1ull };

    if ((-1.0 < x) and (x < 1.0))
    {
        // Includes: subnormal, +0, -0
        // No integral part.
        return x;
    }

    const Bits u                  = std::bit_cast< Bits >(x);
    const Bits exponent_bits      = (u & s_exponent_mask) >> s_mantissa_bit_count;
    assert(s_exponent_bias < exponent_bits);
    const Bits exponent           = exponent_bits - s_exponent_bias;
    if (s_mantissa_bit_count <= exponent)
    {
        // Includes: +Inf, -Inf, NaN
        // No fractional part.
        return {};
    }

    const Bits fraction_bit_count = s_mantissa_bit_count - exponent;
    const Bits fraction_mask      = (1ull << fraction_bit_count) - 1ull;
    const Bits fraction_bits      = u & fraction_mask;
    const Bits fraction_shift     = s_mantissa_bit_count - std::bit_width(fraction_bits)
                                  + 1ull; // Implicit leading one

    Bits fraction = u & s_sign_mask;
    if (fraction_shift < exponent_bits)
    {
        const Bits fraction_exponent = exponent_bits - fraction_shift;
        const Bits fraction_mantissa = (fraction_bits << fraction_shift)
                                     // Remove implicit leading one.
                                     & s_mantissa_mask;

        fraction |= (fraction_exponent << s_mantissa_bit_count);
        fraction |=  fraction_mantissa;
     }

    return std::bit_cast< double >(fraction);
}

根据您自己的偏好，您也可以在 NaN 的情况下返回x 。

测试：

// setprecision
#include <iomanip>
// cout, endl, fixed
#include <iostream>

auto main() -> int
{
    std::cout << std::fixed << std::setprecision(11);
    
    {
        constexpr double d = 7.99999952316;
        constexpr double frac = Frac(d);
        std::cout << "frac(" << d << ") = " << frac << std::endl;
    }

    {
        constexpr double d = 0.5;
        constexpr double frac = Frac(d);
        std::cout << "frac(" << d << ") = " << frac << std::endl;
    }
    {
        constexpr double d = 0.75;
        constexpr double frac = Frac(d);
        std::cout << "frac(" << d << ") = " << frac << std::endl;
    }
    {
        constexpr double d = 6.5;
        constexpr double frac = Frac(d);
        std::cout << "frac(" << d << ") = " << frac << std::endl;
    }
    {
        constexpr double d = 4.711;
        constexpr double frac = Frac(d);
        std::cout << "frac(" << d << ") = " << frac << std::endl;
    }

    return 0;
}

Output

frac(7.99999952316) = 0.99999952316
frac(0.50000000000) = 0.50000000000
frac(0.75000000000) = 0.75000000000
frac(6.50000000000) = 0.50000000000
frac(4.71100000000) = 0.71100000000