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

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.