比较两个双精度浮点数或两个浮点数最有效的方法是什么?

简单地这样做是不正确的:

bool CompareDoubles1 (double A, double B)
{
   return A == B;
}

比如:

bool CompareDoubles2 (double A, double B) 
{
   diff = A - B;
   return (diff < EPSILON) && (-diff < EPSILON);
}

似乎是浪费加工。

有人知道更聪明的浮点比较器吗?


当前回答

我的课程是基于之前发布的答案。非常类似于谷歌的代码,但我使用了一个偏差,将所有NaN值推到0xFF000000以上。这样可以更快地检查NaN。

这段代码是为了演示概念,而不是通用的解决方案。谷歌的代码已经展示了如何计算所有平台特定的值,我不想复制所有这些。我对这段代码做了有限的测试。

typedef unsigned int   U32;
//  Float           Memory          Bias (unsigned)
//  -----           ------          ---------------
//   NaN            0xFFFFFFFF      0xFF800001
//   NaN            0xFF800001      0xFFFFFFFF
//  -Infinity       0xFF800000      0x00000000 ---
//  -3.40282e+038   0xFF7FFFFF      0x00000001    |
//  -1.40130e-045   0x80000001      0x7F7FFFFF    |
//  -0.0            0x80000000      0x7F800000    |--- Valid <= 0xFF000000.
//   0.0            0x00000000      0x7F800000    |    NaN > 0xFF000000
//   1.40130e-045   0x00000001      0x7F800001    |
//   3.40282e+038   0x7F7FFFFF      0xFEFFFFFF    |
//   Infinity       0x7F800000      0xFF000000 ---
//   NaN            0x7F800001      0xFF000001
//   NaN            0x7FFFFFFF      0xFF7FFFFF
//
//   Either value of NaN returns false.
//   -Infinity and +Infinity are not "close".
//   -0 and +0 are equal.
//
class CompareFloat{
public:
    union{
        float     m_f32;
        U32       m_u32;
    };
    static bool   CompareFloat::IsClose( float A, float B, U32 unitsDelta = 4 )
                  {
                      U32    a = CompareFloat::GetBiased( A );
                      U32    b = CompareFloat::GetBiased( B );

                      if ( (a > 0xFF000000) || (b > 0xFF000000) )
                      {
                          return( false );
                      }
                      return( (static_cast<U32>(abs( a - b ))) < unitsDelta );
                  }
    protected:
    static U32    CompareFloat::GetBiased( float f )
                  {
                      U32    r = ((CompareFloat*)&f)->m_u32;

                      if ( r & 0x80000000 )
                      {
                          return( ~r - 0x007FFFFF );
                      }
                      return( r + 0x7F800000 );
                  }
};

其他回答

使用任何其他建议都要非常小心。这完全取决于上下文。

我花了很长时间在一个系统中追踪错误,该系统假设|a-b|<epsilon,则a==b。潜在的问题是:

The implicit presumption in an algorithm that if a==b and b==c then a==c. Using the same epsilon for lines measured in inches and lines measured in mils (.001 inch). That is a==b but 1000a!=1000b. (This is why AlmostEqual2sComplement asks for the epsilon or max ULPS). The use of the same epsilon for both the cosine of angles and the length of lines! Using such a compare function to sort items in a collection. (In this case using the builtin C++ operator == for doubles produced correct results.)

就像我说的,这完全取决于上下文和a和b的预期大小。

顺便说一下,std::numeric_limits<double>::epsilon()是“机器epsilon”。它是1.0和下一个用double表示的值之间的差值。我猜它可以用在比较函数中,但只有当期望值小于1时。(这是对@cdv的回答的回应…)

同样,如果你的int算术是双精度的(这里我们在某些情况下使用双精度来保存int值),你的算术是正确的。例如,4.0/2.0将等同于1.0+1.0。只要你不做导致分数(4.0/3.0)的事情,或者不超出int的大小。

Why not perform bitwise XOR? Two floating point numbers are equal if their corresponding bits are equal. I think, the decision to place the exponent bits before mantissa was made to speed up comparison of two floats. I think, many answers here are missing the point of epsilon comparison. Epsilon value only depends on to what precision floating point numbers are compared. For example, after doing some arithmetic with floats you get two numbers: 2.5642943554342 and 2.5642943554345. They are not equal, but for the solution only 3 decimal digits matter so then they are equal: 2.564 and 2.564. In this case you choose epsilon equal to 0.001. Epsilon comparison is also possible with bitwise XOR. Correct me if I am wrong.

这是另一个解:

#include <cmath>
#include <limits>

auto Compare = [](float a, float b, float epsilon = std::numeric_limits<float>::epsilon()){ return (std::fabs(a - b) <= epsilon); };

就数量的规模而言:

如果在某种物理意义上,ε是量的大小(即相对值)的一小部分,而A和B类型在同一意义上具有可比性,那么我认为,下面的观点是相当正确的:

#include <limits>
#include <iomanip>
#include <iostream>

#include <cmath>
#include <cstdlib>
#include <cassert>

template< typename A, typename B >
inline
bool close_enough(A const & a, B const & b,
                  typename std::common_type< A, B >::type const & epsilon)
{
    using std::isless;
    assert(isless(0, epsilon)); // epsilon is a part of the whole quantity
    assert(isless(epsilon, 1));
    using std::abs;
    auto const delta = abs(a - b);
    auto const x = abs(a);
    auto const y = abs(b);
    // comparable generally and |a - b| < eps * (|a| + |b|) / 2
    return isless(epsilon * y, x) && isless(epsilon * x, y) && isless((delta + delta) / (x + y), epsilon);
}

int main()
{
    std::cout << std::boolalpha << close_enough(0.9, 1.0, 0.1) << std::endl;
    std::cout << std::boolalpha << close_enough(1.0, 1.1, 0.1) << std::endl;
    std::cout << std::boolalpha << close_enough(1.1,    1.2,    0.01) << std::endl;
    std::cout << std::boolalpha << close_enough(1.0001, 1.0002, 0.01) << std::endl;
    std::cout << std::boolalpha << close_enough(1.0, 0.01, 0.1) << std::endl;
    return EXIT_SUCCESS;
}

这取决于你想要的比较有多精确。如果您想对完全相同的数字进行比较,那么只需使用==。(除非你真的想要完全相同的数字,否则你几乎不会想这么做。)在任何一个不错的平台上,你都可以做到以下几点:

diff= a - b; return fabs(diff)<EPSILON;

因为晶圆厂往往很快。我说的快是指它基本上是一个位与,所以它最好快。

用于比较双精度和浮点数的整数技巧很好,但往往会使各种CPU管道更难有效处理。现在,由于使用堆栈作为频繁使用的值的临时存储区域,在某些有序架构上它肯定不会更快。(在乎的人可以去Load-hit-store。)