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

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.

其他回答

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 );
                  }
};

在c++中获取的可移植方法是

#include <limits>
std::numeric_limits<double>::epsilon()

然后比较函数变成

#include <cmath>
#include <limits>

bool AreSame(double a, double b) {
    return std::fabs(a - b) < std::numeric_limits<double>::epsilon();
}
/// testing whether two doubles are almost equal. We consider two doubles
/// equal if the difference is within the range [0, epsilon).
///
/// epsilon: a positive number (supposed to be small)
///
/// if either x or y is 0, then we are comparing the absolute difference to
/// epsilon.
/// if both x and y are non-zero, then we are comparing the relative difference
/// to epsilon.
bool almost_equal(double x, double y, double epsilon)
{
    double diff = x - y;
    if (x != 0 && y != 0){
        diff = diff/y; 
    }

    if (diff < epsilon && -1.0*diff < epsilon){
        return true;
    }
    return false;
}

我在我的小项目中使用了这个函数,它是有效的,但注意以下几点:

双精度误差可以为你制造惊喜。假设epsilon = 1.0e-6,那么根据上面的代码,1.0和1.000001不应该被认为是相等的,但在我的机器上,函数认为它们是相等的,这是因为1.000001不能精确地转换为二进制格式,它可能是1.0000009xxx。我用1.0和1.0000011测试了它,这次我得到了预期的结果。

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。)

与epsilon值进行比较是大多数人所做的(甚至是在游戏编程中)。

你应该稍微改变你的实现:

bool AreSame(double a, double b)
{
    return fabs(a - b) < EPSILON;
}

编辑:克里斯特在最近的一篇博客文章中添加了一堆关于这个主题的很棒的信息。享受。