问题描述
我正在使用双精度求解方程，并且得到了-7.07649e-17而不是0的解决方案.
I was solving an equation using double precision and I got -7.07649e-17 as a solution instead of 0.
我同意它已经足够接近，可以说它相等了，但是我已经阅读了该机器C ++双重类型的epsilon是2^-52 ，它大于我得到的值.
I agree it's close enough that I can say it's equal but I've read that the machine epsilon for the C++ double type is 2^-52 which is larger than the value I get.
那么为什么我的价值不如机器epsilon?为什么值不能四舍五入?
So why do I have an inferior value than the machine epsilon?Why isn't the value rounded to zero?
这没什么大不了的，但是当我进行逻辑测试时，我的值似乎不为零...
It's not a big deal but when I do a logical test it appears that my value is not zero...
推荐答案
这个故事有两个不同的常量.一个是epsilon，这是一个最小值，当添加到1.0时会产生不同于1.0的值.如果将较小的值添加到1.0，您将再次得到1.0，因为计算机中数字表示存在物理限制.但是有些值小于epsilon但大于零.对于通过std::numeric_limits<double>::min获得的double，此类数字最小.
There are two different constants in this story. One is epsilon, which is a minimal value that when added to 1.0 produces a value different from 1.0. If you add a smaller value to 1.0 you will again get a 1.0, because there are physical limits to the representation of a number in a computer. But there are values that are less than epsilon and greater than zero. Smallest such number for a double you get with std::numeric_limits<double>::min.
作为参考，您将获得带有std::numeric_limits<double>::epsilon的epsilon.
For reference, you get epsilon with std::numeric_limits<double>::epsilon.
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