float numerical error. How does c++ know 0.99999982 is 1?

Question

I normalized a vector and length of it should be 1. But the result of length method is 0.99999982

I don't know it's right or wrong. But if I print it out, the result is 1. Not 0.99999982( printed by cout )

But how std::cout knows it's 1? [This is my first question]

And another question is why the result of comparing function is false.

I have a comparing method like below. And lhs is the length of a vector and the rhs is just 1.

return (fabsf(rhs-lhs) <= FLT_EPSILON) ? true : false;

Result of this method is false.

Is length of the normalized vector is already wrong to be considered normalized? or epsilon is too small?

What did I wrong?

I like this answer: http://stackoverflow.com/questions/17333/most-effective-way-for-float-and-double-comparison/77735#77735 — R. Martinho Fernandes, Sep 21 '11 at 04:05
Note: return (fabsf(rhs-lhs) <= FLT_EPSILON) ? true : false; should be simplified to return (fabsf(rhs-lhs) <= FLT_EPSILON); — arsenm, Sep 21 '11 at 04:34

score 4 · Answer 1 · answered Sep 21 '11 at 03:33

4

The epsilon is too small. That's because epsilon is the smallest as it can be. (by definition)

So you will need a larger tolerance.

answered Sep 21 '11 at 03:33

Mysticial

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Right. `FLT_EPSILON` is absolutely not useful for a fuzzy equality test. The confusion comes because the symbol epsilon is often used in such formulas, but it's a different epsilon. – Ben Voigt Sep 21 '11 at 03:35

score 1 · Accepted Answer · answered Sep 21 '11 at 04:29

The IEEE standard requires that the result of addition, subtraction, multiplication and division be exactly rounded. The result must then be rounded to the nearest floating-point number. Floating point error is accumulated in the many operators it takes to normalize a vector. Like the other poster said FLT_EPSILON is too small.

float numerical error. How does c++ know 0.99999982 is 1?

2 Answers2