今天，我在看一些c++代码(别人写的)，发现了这一部分:
double someValue = ...
if (someValue < std::numeric_limits<double>::epsilon() &&
someValue > -std::numeric_limits<double>::epsilon()) {
someValue = 0.0;
}
我在想这到底说得通不合理。
epsilon()的文档说:
该函数返回1与可[用双精度符号]表示的大于1的最小值之间的差值。
这是否也适用于0，即()的最小值大于0?或者有没有0到0 +之间的数可以用双精度数表示?
如果不是，那么比较是不是等同于someValue == 0.0?
当前回答
使用IEEE浮点，在最小的非零正数和最小的非零负数之间，存在两个值:正零和负零。测试一个值是否在最小的非零值之间等价于测试与零相等;然而，赋值可能会产生影响，因为它会将负0变为正0。
It would be conceivable that a floating-point format might have three values between the smallest finite positive and negative values: positive infinitesimal, unsigned zero, and negative infinitesimal. I am not familiar with any floating-point formats that in fact work that way, but such a behavior would be perfectly reasonable and arguably better than that of IEEE (perhaps not enough better to be worth adding extra hardware to support it, but mathematically 1/(1/INF), 1/(-1/INF), and 1/(1-1) should represent three distinct cases illustrating three different zeroes). I don't know whether any C standard would mandate that signed infinitesimals, if they exist, would have to compare equal to zero. If they do not, code like the above could usefully ensure that e.g. dividing a number repeatedly by two would eventually yield zero rather than being stuck on "infinitesimal".
其他回答
该测试当然与someValue == 0不同。浮点数的全部思想是存储一个指数和一个显著值。因此，它们表示具有一定数量的精度二进制有效位数的值(在IEEE双精度的情况下为53)。可表示值在0附近比在1附近密集得多。
为了使用更熟悉的十进制系统，假设您使用exponent存储一个“4位有效数字”的十进制值。那么下一个大于1的可表示值是1.001 * 10^0，是1.000 * 10^ 3。但是1.000 * 10^-4也是可以表示的，假设指数可以存储-4。你可以相信我的话，IEEE double可以存储小于的指数。
你不能仅仅从这段代码中判断用作为边界是否有意义，你需要看一下上下文。可能是对产生someValue的计算错误的合理估计，也可能不是。
使用IEEE浮点，在最小的非零正数和最小的非零负数之间，存在两个值:正零和负零。测试一个值是否在最小的非零值之间等价于测试与零相等;然而，赋值可能会产生影响，因为它会将负0变为正0。
It would be conceivable that a floating-point format might have three values between the smallest finite positive and negative values: positive infinitesimal, unsigned zero, and negative infinitesimal. I am not familiar with any floating-point formats that in fact work that way, but such a behavior would be perfectly reasonable and arguably better than that of IEEE (perhaps not enough better to be worth adding extra hardware to support it, but mathematically 1/(1/INF), 1/(-1/INF), and 1/(1-1) should represent three distinct cases illustrating three different zeroes). I don't know whether any C standard would mandate that signed infinitesimals, if they exist, would have to compare equal to zero. If they do not, code like the above could usefully ensure that e.g. dividing a number repeatedly by two would eventually yield zero rather than being stuck on "infinitesimal".
你不能把这个应用到0，因为有尾数和指数部分。 由于指数可以存储很小的数，小于， 但是当你尝试做一些类似(1.0 -“非常小的数字”)的事情时，你会得到1.0。 Epsilon不是值的指示器，而是值精度的指示器，值精度是尾数。 它显示了我们可以存储多少个正确的十进制数字。
假设64位IEEE双精度，则有52位尾数和11位指数。让我们把它分解一下:
1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^0 = 1
大于1的最小可表示数:
1.0000 00000000 00000000 00000000 00000000 00000000 00000001 × 2^0 = 1 + 2^-52
因此:
epsilon = (1 + 2^-52) - 1 = 2^-52
在0和之间有数字吗?很多……例如，最小正可表示(正常)数为:
1.0000 00000000 00000000 00000000 00000000 00000000 00000000 × 2^-1022 = 2^-1022
事实上，在0和之间有(1022 - 52 + 1)×2^52 = 4372995238176751616个数字，这是所有正可表示数字的47%…
假设系统无法区分1.000000000000000000000和1.00000000000000001。这是1.0和1.0 + 1e-20。你认为在-1e-20和+1e-20之间还有一些值可以表示吗?
