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?
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 );
}
};
我为java编写这篇文章,但是您可能会发现它很有用。它使用长变量而不是双变量,但会处理nan、亚法线等。
public static boolean equal(double a, double b) {
final long fm = 0xFFFFFFFFFFFFFL; // fraction mask
final long sm = 0x8000000000000000L; // sign mask
final long cm = 0x8000000000000L; // most significant decimal bit mask
long c = Double.doubleToLongBits(a), d = Double.doubleToLongBits(b);
int ea = (int) (c >> 52 & 2047), eb = (int) (d >> 52 & 2047);
if (ea == 2047 && (c & fm) != 0 || eb == 2047 && (d & fm) != 0) return false; // NaN
if (c == d) return true; // identical - fast check
if (ea == 0 && eb == 0) return true; // ±0 or subnormals
if ((c & sm) != (d & sm)) return false; // different signs
if (abs(ea - eb) > 1) return false; // b > 2*a or a > 2*b
d <<= 12; c <<= 12;
if (ea < eb) c = c >> 1 | sm;
else if (ea > eb) d = d >> 1 | sm;
c -= d;
return c < 65536 && c > -65536; // don't use abs(), because:
// There is a posibility c=0x8000000000000000 which cannot be converted to positive
}
public static boolean zero(double a) { return (Double.doubleToLongBits(a) >> 52 & 2047) < 3; }
请记住,在一些浮点运算之后,number可能与我们期望的非常不同。没有代码可以解决这个问题。
/// 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。
正如其他人所指出的那样,使用固定指数(例如0.0000001)对于远离该值的值是无用的。例如,如果你的两个值是10000.000977和10000,那么这两个数字之间没有32位浮点值——10000和10000.000977是你可能得到的最接近的值,而不是位对位相同。这里,小于0.0009是没有意义的;你也可以使用直接等式运算符。
同样地,当两个值的大小接近ε时,相对误差增长到100%。
Thus, trying to mix a fixed point number such as 0.00001 with floating-point values (where the exponent is arbitrary) is a pointless exercise. This will only ever work if you can be assured that the operand values lie within a narrow domain (that is, close to some specific exponent), and if you properly select an epsilon value for that specific test. If you pull a number out of the air ("Hey! 0.00001 is small, so that must be good!"), you're doomed to numerical errors. I've spent plenty of time debugging bad numerical code where some poor schmuck tosses in random epsilon values to make yet another test case work.
如果你从事任何类型的数值编程,并认为你需要达到定点的epsilon,请阅读BRUCE关于比较浮点数的文章。
浮点数比较