Question

I was doing some rounding calculations and happened upon a question. How can I express the highest quantity less than 1 for a given floating point type?

That is, how I write/represent value x such that x < 1, x + y >= 1 for any y > 0.

In fractions this would be x = (q-1)/q where q is the precision of the type. For example, if you are counting in 1/999 increments then x = 998/999.

For a given type (float, double, long double), how could one express the value x in code?

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I also wonder if such a value actually exists for all values of y. That is, as y's exponent gets smaller perhaps the relation doesn’t hold anymore. So an answer with some range restriction on y is also acceptable. (The value of x I want still does exist, the relationship may just not properly express it.)

Answer 1

C99 defines nextafter() function. Use it like

#include <math.h>
double under_one = nextafter(1, 0);