Java double losing precision calculation

Question

I have the following double calculation in Java:

double C1 = some value;    // It's always an integer
double C2 = some value;    // It's always integer, and C2 >= C1
double S = some value;     // It's always 0 < S <= 1
double carry = some value; // It's always 0 <= carry < 1

double rate = ((C1/C2)/S);
double C3 = Math.floor( (rate * S * C2) + carry );

Now, if we did not lose precision, C3 == C1 would be true. But since we lose precision, will C3 and C1 still always be equal?

If they would not always be equal, if I have C1, C2, S, and C3, how could I modify the rate to make sure after calculation C3 would be equal to C1?

Note: unfortunately using BigDecimal is not an option.

Answer 1

Generally, we don't compare two double values directly through ==. Instead, we measure the difference between them. e.g. Math.abs(C3 - C1) < 1e-6.

Answer 2

Not so.

Floating point is an approximation. The next double, bit 1 added to the mantissa can cause a quite large gap for large exponents between those two doubles.

S approximating 0 will cause huge doubles, that have a large gap between them. That will no longer be correctable, say with that simple rounding. Even if the small S does not cause the division to exceed the double range with INFINITY.

One can easily be of by 1000.