问题
I was solving an equation using double precision and I got -7.07649e-17 as a solution instead of 0.
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.
So why do I have an inferior value than the machine epsilon? Why isn't the value rounded to zero?
It's not a big deal but when I do a logical test it appears that my value is not zero...
回答1:
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.
For reference, you get epsilon with std::numeric_limits<double>::epsilon.
回答2:
You are not guaranteed that rounding will take place at any particular time. The C++ standard permits the implementation to use additional precision pretty much anywhere it wants to and many real-world implementations do exactly that.
回答3:
A common solution for the floating point precision problem is to define an epsilon value yourself and compare to that instead of zero.
e.g.
double epsilon = 0.00001;
if (abs(value) < epsilon) // treat value as 0 in your code
来源：https://stackoverflow.com/questions/43678480/why-my-double-can-contain-a-value-below-the-machine-epsilon