whats the biggest number that is smaller than FLT_EPSILON?

Question

In C using float.h I wonder how can I find the largest number that if I add to 1 the answer will remain 1.

ie 1 + x = 1 how to find x?

Answer 1

If you want "the biggest number that is smaller than FLT_EPSILON":

x = nextafterf(FLT_EPSILON, 0)

If you want the biggest number x such that 1.0f + x == 1.0f , then the answer depends on the rounding mode, but in the default rounding mode, it's simply FLT_EPSILON/2 .

However, things aren't quite so simple. Because of the way round-to-nearest rounds to the even neighbor on ties, you have 1.0f + FLT_EPSILON/2 == 1.0f , but:

(1.0f+FLT_EPSILON) + FLT_EPSILON/2 != (1.0f+FLT_EPSILON)

So you may instead want to use the slightly smaller value of x :

x = nextafterf(FLT_EPSILON/2, 0)

This will ensure that y+x == y for any y >= 1.0 .

Answer 2

It depends on rounding modes. Here's a simple example. Suppose our precision is 4 bits, and we have some sort of IEEE754 representation. So the value 1 is stored as 1.0000 × 2 ⁰ . The next bigger number is 1.0001 × 2 ⁰ , and the machine epsilon ε is defined as the difference between the two, which is 0.0001 × 2 ⁰ = 1.0000 × 2 ⁻⁴ . Now:

• ε / 2 = 1.0000 × 2 ⁻⁵ = 0.00001, and

• ε / 4 = 1.0000 × 2 ⁻⁶ = 0.000001.

When you add one of the two to 1, the numbers are first rewritten for the power 2 ⁰ and rounded to 4 digits after the separator. The surviving mantissa of ε / 4 is surely 0.0000, while the surviving mantissa of ε / 2 is either 0.0000 or 0.0001, depending on whether you round up or down.

As long as the effective resulting mantissa is 0.0000, you can add the number to 1 without changing its value.

(The actual precisions are 23+1, 52+1 and 64 for single, double and extended double precision floats.)