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

你必须为浮点数比较做这个处理,因为浮点数不能像整数类型那样完美地比较。下面是各种比较运算符的函数。

浮点数等于(==)

我也更喜欢减法技术,而不是依赖于fabs()或abs(),但我必须在从64位PC到ATMega328微控制器(Arduino)的各种架构上快速配置它,才能真正看到它是否会产生很大的性能差异。

所以,让我们忘记这些绝对值的东西,只做一些减法和比较!

从微软的例子修改如下:

/// @brief      See if two floating point numbers are approximately equal.
/// @param[in]  a        number 1
/// @param[in]  b        number 2
/// @param[in]  epsilon  A small value such that if the difference between the two numbers is
///                      smaller than this they can safely be considered to be equal.
/// @return     true if the two numbers are approximately equal, and false otherwise
bool is_float_eq(float a, float b, float epsilon) {
    return ((a - b) < epsilon) && ((b - a) < epsilon);
}
bool is_double_eq(double a, double b, double epsilon) {
    return ((a - b) < epsilon) && ((b - a) < epsilon);
}

使用示例:

constexpr float EPSILON = 0.0001; // 1e-4
is_float_eq(1.0001, 0.99998, EPSILON);

我不完全确定,但在我看来,对基于epsilon的方法的一些批评,正如这个高度好评的答案下面的评论所描述的那样,可以通过使用变量epsilon来解决,根据比较的浮点值缩放,像这样:

float a = 1.0001;
float b = 0.99998;
float epsilon = std::max(std::fabs(a), std::fabs(b)) * 1e-4;

is_float_eq(a, b, epsilon);

通过这种方式,epsilon值随浮点值伸缩,因此它的值不会小到不重要。

为了完整起见,让我们添加剩下的:

大于(>)小于(<):

/// @brief      See if floating point number `a` is > `b`
/// @param[in]  a        number 1
/// @param[in]  b        number 2
/// @param[in]  epsilon  a small value such that if `a` is > `b` by this amount, `a` is considered
///             to be definitively > `b`
/// @return     true if `a` is definitively > `b`, and false otherwise
bool is_float_gt(float a, float b, float epsilon) {
    return a > b + epsilon;
}
bool is_double_gt(double a, double b, double epsilon) {
    return a > b + epsilon;
}

/// @brief      See if floating point number `a` is < `b`
/// @param[in]  a        number 1
/// @param[in]  b        number 2
/// @param[in]  epsilon  a small value such that if `a` is < `b` by this amount, `a` is considered
///             to be definitively < `b`
/// @return     true if `a` is definitively < `b`, and false otherwise
bool is_float_lt(float a, float b, float epsilon) {
    return a < b - epsilon;
}
bool is_double_lt(double a, double b, double epsilon) {
    return a < b - epsilon;
}

大于或等于(>=),小于或等于(<=)

/// @brief      Returns true if `a` is definitively >= `b`, and false otherwise
bool is_float_ge(float a, float b, float epsilon) {
    return a > b - epsilon;
}
bool is_double_ge(double a, double b, double epsilon) {
    return a > b - epsilon;
}

/// @brief      Returns true if `a` is definitively <= `b`, and false otherwise
bool is_float_le(float a, float b, float epsilon) {
    return a < b + epsilon;
}
bool is_double_le(double a, double b, double epsilon) {
    return a < b + epsilon;
}

额外的改进:

A good default value for epsilon in C++ is std::numeric_limits<T>::epsilon(), which evaluates to either 0 or FLT_EPSILON, DBL_EPSILON, or LDBL_EPSILON. See here: https://en.cppreference.com/w/cpp/types/numeric_limits/epsilon. You can also see the float.h header for FLT_EPSILON, DBL_EPSILON, and LDBL_EPSILON. See https://en.cppreference.com/w/cpp/header/cfloat and https://www.cplusplus.com/reference/cfloat/ You could template the functions instead, to handle all floating point types: float, double, and long double, with type checks for these types via a static_assert() inside the template. Scaling the epsilon value is a good idea to ensure it works for really large and really small a and b values. This article recommends and explains it: http://realtimecollisiondetection.net/blog/?p=89. So, you should scale epsilon by a scaling value equal to max(1.0, abs(a), abs(b)), as that article explains. Otherwise, as a and/or b increase in magnitude, the epsilon would eventually become so small relative to those values that it becomes lost in the floating point error. So, we scale it to become larger in magnitude like they are. However, using 1.0 as the smallest allowed scaling factor for epsilon also ensures that for really small-magnitude a and b values, epsilon itself doesn't get scaled so small that it also becomes lost in the floating point error. So, we limit the minimum scaling factor to 1.0. If you want to "encapsulate" the above functions into a class, don't. Instead, wrap them up in a namespace if you like in order to namespace them. Ex: if you put all of the stand-alone functions into a namespace called float_comparison, then you could access the is_eq() function like this, for instance: float_comparison::is_eq(1.0, 1.5);. It might also be nice to add comparisons against zero, not just comparisons between two values. So, here is a better type of solution with the above improvements in place: namespace float_comparison { /// Scale the epsilon value to become large for large-magnitude a or b, /// but no smaller than 1.0, per the explanation above, to ensure that /// epsilon doesn't ever fall out in floating point error as a and/or b /// increase in magnitude. template<typename T> static constexpr T scale_epsilon(T a, T b, T epsilon = std::numeric_limits<T>::epsilon()) noexcept { static_assert(std::is_floating_point_v<T>, "Floating point comparisons " "require type float, double, or long double."); T scaling_factor; // Special case for when a or b is infinity if (std::isinf(a) || std::isinf(b)) { scaling_factor = 0; } else { scaling_factor = std::max({(T)1.0, std::abs(a), std::abs(b)}); } T epsilon_scaled = scaling_factor * std::abs(epsilon); return epsilon_scaled; } // Compare two values /// Equal: returns true if a is approximately == b, and false otherwise template<typename T> static constexpr bool is_eq(T a, T b, T epsilon = std::numeric_limits<T>::epsilon()) noexcept { static_assert(std::is_floating_point_v<T>, "Floating point comparisons " "require type float, double, or long double."); // test `a == b` first to see if both a and b are either infinity // or -infinity return a == b || std::abs(a - b) <= scale_epsilon(a, b, epsilon); } /* etc. etc.: is_eq() is_ne() is_lt() is_le() is_gt() is_ge() */ // Compare against zero /// Equal: returns true if a is approximately == 0, and false otherwise template<typename T> static constexpr bool is_eq_zero(T a, T epsilon = std::numeric_limits<T>::epsilon()) noexcept { static_assert(std::is_floating_point_v<T>, "Floating point comparisons " "require type float, double, or long double."); return is_eq(a, (T)0.0, epsilon); } /* etc. etc.: is_eq_zero() is_ne_zero() is_lt_zero() is_le_zero() is_gt_zero() is_ge_zero() */ } // namespace float_comparison

参见:

The macro forms of some of the functions above in my repo here: utilities.h. UPDATE 29 NOV 2020: it's a work-in-progress, and I'm going to make it a separate answer when ready, but I've produced a better, scaled-epsilon version of all of the functions in C in this file here: utilities.c. Take a look. ADDITIONAL READING I need to do now have done: Floating-point tolerances revisited, by Christer Ericson. VERY USEFUL ARTICLE! It talks about scaling epsilon in order to ensure it never falls out in floating point error, even for really large-magnitude a and/or b values!

其他回答

在这个版本中,你可以检查,这些数字之间的差异并不比某些分数(比如,0.0001%)更大:

bool floatApproximatelyEquals(const float a, const float b) {
    if (b == 0.) return a == 0.; // preventing division by zero
    return abs(1. - a / b) < 1e-6;
}

请注意Sneftel关于浮动可能的分数限制的评论。

还要注意的是,它不同于使用绝对的epsilon的方法——这里你不需要担心“数量级”——数字可能是,比如说1e100,或者1e-100,它们总是会被一致地比较,而且你不必为每一种情况更新epsilon。

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.

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。

有关更深入的方法,请参阅比较浮点数。以下是该链接的代码片段:

// Usable AlmostEqual function    
bool AlmostEqual2sComplement(float A, float B, int maxUlps)    
{    
    // Make sure maxUlps is non-negative and small enough that the    
    // default NAN won't compare as equal to anything.    
    assert(maxUlps > 0 && maxUlps < 4 * 1024 * 1024);    
    int aInt = *(int*)&A;    
    // Make aInt lexicographically ordered as a twos-complement int    
    if (aInt < 0)    
        aInt = 0x80000000 - aInt;    
    // Make bInt lexicographically ordered as a twos-complement int    
    int bInt = *(int*)&B;    
    if (bInt < 0)    
        bInt = 0x80000000 - bInt;    
    int intDiff = abs(aInt - bInt);    
    if (intDiff <= maxUlps)    
        return true;    
    return false;    
}

你写的代码有bug:

return (diff < EPSILON) && (-diff > EPSILON);

正确的代码应该是:

return (diff < EPSILON) && (diff > -EPSILON);

(…是的,这是不同的)

我想知道晶圆厂是否会让你在某些情况下失去懒惰的评价。我会说这取决于编译器。你可能想两种都试试。如果它们在平均水平上是相等的,则采用晶圆厂实现。

如果你有一些关于两个浮点数中哪一个比另一个更大的信息,你可以根据比较的顺序来更好地利用惰性求值。

最后,通过内联这个函数可能会得到更好的结果。不过不太可能有太大改善……

编辑:OJ,谢谢你纠正你的代码。我相应地删除了我的评论