Why my double can contain a value below the machine epsilon?

Question

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 s

Answer 1

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

Answer 2

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.

Answer 3

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.