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。

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。

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.

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();
}

Qt实现了两个函数，也许你可以从中学到一些东西:

static inline bool qFuzzyCompare(double p1, double p2)
{
    return (qAbs(p1 - p2) <= 0.000000000001 * qMin(qAbs(p1), qAbs(p2)));
}

static inline bool qFuzzyCompare(float p1, float p2)
{
    return (qAbs(p1 - p2) <= 0.00001f * qMin(qAbs(p1), qAbs(p2)));
}

您可能需要以下函数，因为

请注意，比较p1或p2为0.0的值是无效的， 也不会比较其中一个值为NaN或无穷大的值。 如果其中一个值总是0.0，则使用qFuzzyIsNull代替。如果一个人 其中的值很可能是0.0，一种解决方案是将两者都加上1.0 值。

static inline bool qFuzzyIsNull(double d)
{
    return qAbs(d) <= 0.000000000001;
}

static inline bool qFuzzyIsNull(float f)
{
    return qAbs(f) <= 0.00001f;
}

比较浮点数取决于上下文。因为即使改变操作的顺序也会产生不同的结果，所以知道你希望这些数字有多“相等”是很重要的。

在研究浮点数比较时，比较Bruce Dawson编写的浮点数是一个很好的开始。

以下定义来自Knuth的《The art of computer programming》:

bool approximatelyEqual(float a, float b, float epsilon)
{
    return fabs(a - b) <= ( (fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}

bool essentiallyEqual(float a, float b, float epsilon)
{
    return fabs(a - b) <= ( (fabs(a) > fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}

bool definitelyGreaterThan(float a, float b, float epsilon)
{
    return (a - b) > ( (fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}

bool definitelyLessThan(float a, float b, float epsilon)
{
    return (b - a) > ( (fabs(a) < fabs(b) ? fabs(b) : fabs(a)) * epsilon);
}

当然，选择取决于上下文，并决定你想要的数字有多相等。

比较浮点数的另一种方法是查看数字的ULP(最后位置的单位)。虽然没有专门处理比较，但“每个计算机科学家都应该知道浮点数”这篇论文是了解浮点数如何工作以及陷阱是什么，包括什么是ULP的很好的资源。