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
我使用以下函数进行浮点数比较:
bool approximatelyEqual(double a, double b)
{
return fabs(a - b) <= ((fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * std::numeric_limits<double>::epsilon());
}
其他回答
下面是使用std::numeric_limits::epsilon()不是答案的证明——对于大于1的值它会失败:
证明我上面的评论:
#include <stdio.h>
#include <limits>
double ItoD (__int64 x) {
// Return double from 64-bit hexadecimal representation.
return *(reinterpret_cast<double*>(&x));
}
void test (__int64 ai, __int64 bi) {
double a = ItoD(ai), b = ItoD(bi);
bool close = std::fabs(a-b) < std::numeric_limits<double>::epsilon();
printf ("%.16f and %.16f %s close.\n", a, b, close ? "are " : "are not");
}
int main()
{
test (0x3fe0000000000000L,
0x3fe0000000000001L);
test (0x3ff0000000000000L,
0x3ff0000000000001L);
}
运行产生如下输出:
0.5000000000000000 and 0.5000000000000001 are close.
1.0000000000000000 and 1.0000000000000002 are not close.
请注意，在第二种情况下(1且仅大于1)，两个输入值尽可能接近，并且仍然比较为不接近。因此，对于大于1.0的值，不妨只使用相等性测试。当比较浮点值时，固定的epsilon将无法保存您的数据。
General-purpose comparison of floating-point numbers is generally meaningless. How to compare really depends on a problem at hand. In many problems, numbers are sufficiently discretized to allow comparing them within a given tolerance. Unfortunately, there are just as many problems, where such trick doesn't really work. For one example, consider working with a Heaviside (step) function of a number in question (digital stock options come to mind) when your observations are very close to the barrier. Performing tolerance-based comparison wouldn't do much good, as it would effectively shift the issue from the original barrier to two new ones. Again, there is no general-purpose solution for such problems and the particular solution might require going as far as changing the numerical method in order to achieve stability.
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。
我使用这个代码:
bool AlmostEqual(double v1, double v2)
{
return (std::fabs(v1 - v2) < std::fabs(std::min(v1, v2)) * std::numeric_limits<double>::epsilon());
}
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。)
