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

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。)

其他回答

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。

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。

我最终花了相当多的时间在这个伟大的线程通过材料。我怀疑每个人都想花这么多时间,所以我将强调我所学到的总结和我实施的解决方案。

快速的总结

Is 1e-8 approximately same as 1e-16? If you are looking at noisy sensor data then probably yes but if you are doing molecular simulation then may be not! Bottom line: You always need to think of tolerance value in context of specific function call and not just make it generic app-wide hard-coded constant. For general library functions, it's still nice to have parameter with default tolerance. A typical choice is numeric_limits::epsilon() which is same as FLT_EPSILON in float.h. This is however problematic because epsilon for comparing values like 1.0 is not same as epsilon for values like 1E9. The FLT_EPSILON is defined for 1.0. The obvious implementation to check if number is within tolerance is fabs(a-b) <= epsilon however this doesn't work because default epsilon is defined for 1.0. We need to scale epsilon up or down in terms of a and b. There are two solution to this problem: either you set epsilon proportional to max(a,b) or you can get next representable numbers around a and then see if b falls into that range. The former is called "relative" method and later is called ULP method. Both methods actually fails anyway when comparing with 0. In this case, application must supply correct tolerance.

实用函数实现(c++ 11)

//implements relative method - do not use for comparing with zero
//use this most of the time, tolerance needs to be meaningful in your context
template<typename TReal>
static bool isApproximatelyEqual(TReal a, TReal b, TReal tolerance = std::numeric_limits<TReal>::epsilon())
{
    TReal diff = std::fabs(a - b);
    if (diff <= tolerance)
        return true;

    if (diff < std::fmax(std::fabs(a), std::fabs(b)) * tolerance)
        return true;

    return false;
}

//supply tolerance that is meaningful in your context
//for example, default tolerance may not work if you are comparing double with float
template<typename TReal>
static bool isApproximatelyZero(TReal a, TReal tolerance = std::numeric_limits<TReal>::epsilon())
{
    if (std::fabs(a) <= tolerance)
        return true;
    return false;
}


//use this when you want to be on safe side
//for example, don't start rover unless signal is above 1
template<typename TReal>
static bool isDefinitelyLessThan(TReal a, TReal b, TReal tolerance = std::numeric_limits<TReal>::epsilon())
{
    TReal diff = a - b;
    if (diff < tolerance)
        return true;

    if (diff < std::fmax(std::fabs(a), std::fabs(b)) * tolerance)
        return true;

    return false;
}
template<typename TReal>
static bool isDefinitelyGreaterThan(TReal a, TReal b, TReal tolerance = std::numeric_limits<TReal>::epsilon())
{
    TReal diff = a - b;
    if (diff > tolerance)
        return true;

    if (diff > std::fmax(std::fabs(a), std::fabs(b)) * tolerance)
        return true;

    return false;
}

//implements ULP method
//use this when you are only concerned about floating point precision issue
//for example, if you want to see if a is 1.0 by checking if its within
//10 closest representable floating point numbers around 1.0.
template<typename TReal>
static bool isWithinPrecisionInterval(TReal a, TReal b, unsigned int interval_size = 1)
{
    TReal min_a = a - (a - std::nextafter(a, std::numeric_limits<TReal>::lowest())) * interval_size;
    TReal max_a = a + (std::nextafter(a, std::numeric_limits<TReal>::max()) - a) * interval_size;

    return min_a <= b && max_a >= b;
}

在数值软件中,确实有这样的情况,你需要检查两个浮点数是否完全相等。我就一个类似的问题发表了这篇文章

https://stackoverflow.com/a/10973098/1447411

所以你不能说“CompareDoubles1”是错误的。

在https://en.cppreference.com/w/cpp/types/numeric_limits/epsilon上找到了另一个有趣的实现

#include <cmath>
#include <limits>
#include <iomanip>
#include <iostream>
#include <type_traits>
#include <algorithm>



template<class T>
typename std::enable_if<!std::numeric_limits<T>::is_integer, bool>::type
    almost_equal(T x, T y, int ulp)
{
    // the machine epsilon has to be scaled to the magnitude of the values used
    // and multiplied by the desired precision in ULPs (units in the last place)
    return std::fabs(x-y) <= std::numeric_limits<T>::epsilon() * std::fabs(x+y) * ulp
        // unless the result is subnormal
        || std::fabs(x-y) < std::numeric_limits<T>::min();
}

int main()
{
    double d1 = 0.2;
    double d2 = 1 / std::sqrt(5) / std::sqrt(5);
    std::cout << std::fixed << std::setprecision(20) 
        << "d1=" << d1 << "\nd2=" << d2 << '\n';

    if(d1 == d2)
        std::cout << "d1 == d2\n";
    else
        std::cout << "d1 != d2\n";

    if(almost_equal(d1, d2, 2))
        std::cout << "d1 almost equals d2\n";
    else
        std::cout << "d1 does not almost equal d2\n";
}