3 
auto a{ 4.2 };
auto b{ 0.12 };
auto result = std::fmod(a, b);
if(result <= std::numeric_limits<double>::epsilon())
  result = 0; // <-- This line isn't triggered

In this example, 4.2 is actually equal to 4.2000000000000002 due to double imprecision.

Note that 4.2/0.12 = 35.

I would expect the output to be equal to std::numeric_limits<double>::epsilon(). Instead, result is equal to 1.5 * std::numeric_limits<double>::epsilon().

Where does this 1.5 multiplier come from?

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  • 1
    Why would you expect the result to be equal to exactly epsilon? To verify why it is 1.5*epsilon, you can bring 4.2 and 0.12 into their binary form and then calculate the remainder.
    – chtz
    Aug 20, 2020 at 7:31
  • 3
    The size of the steps between the double numbers in [1, 2) is epsilon. The size of the steps between the double numbers in [4, 8), in which 4.2 lies, is 4*epsilon. The size of the steps between the double numbers in [.0625, .125), in which .12 lies, is epsilon/16. Let’s call these steps, epsilon/16, s. The double nearest 4.2 is apparently 24*s away from the nearest multiple of the double nearest .12. 24*s is 1.5*epsilon. That is where the 1.5 comes from.
    – Eric Postpischil
    Aug 20, 2020 at 9:27

1 Answer 1

Reset to default
-2

The result of std::fmod may be expected to be accurate to within a ULP or so, but the machine epsilon is the ULP of 1, not of the result of any given operation.

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  • 2
    fmod is one of the few floating point operations that always should be exact (assuming all inputs are finite and the divisor is not zero).
    – chtz
    Aug 20, 2020 at 7:32
  • @chtz Yes, I chose that wording because the operation being exact or not is not important to understand one cannot expect results to always be less than the machine epsilon.
    – Acorn
    Aug 20, 2020 at 16:26
  • To say a floating-point operation is exact is to say it has no error; the produced result is exactly the real-arithmetic result. So, yes, it is important to understand expectations about the result: The error will be less than the machine epsilon; it will be zero.
    – Eric Postpischil
    Aug 20, 2020 at 20:05
  • Again, you are discussing guarantees on a particular operation and I am discussing the mistake of relating errors directly to the machine epsilon as if it was some kind of constant delta between the representable points.
    – Acorn
    Aug 21, 2020 at 4:24

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