Distance between doubles, etc

[janzon]

Hello!

Suppose x is a double not equal to Inf or NaN. How can I find the
smallest double y, such that x<y? It would be nice not having to deal
with bits, big and little endian and so forth. (But maybe I have to.)

Is there a quick and dirty way to get a rough estimate? Is the number
DBL_EPSILON*x of any value?

 

[Pete Becker]

Suppose x is a double not equal to Inf or NaN. How can I find the
smallest double y, such that x<y? It would be nice not having to deal
with bits, big and little endian and so forth. (But maybe I have to.)

nextafter(x, numeric_limits<double>::max())

nextafter was added to C in C99, and is in C++'s TR1. It's also been
voted into the next version of the standard.

If your implementation doesn't supply it you'll probably have to do some
bit twiddling.

--

-- Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com)
Author of "The Standard C++ Library Extensions: a Tutorial and
Reference." (www.petebecker.com/tr1book)

 

[loic-dev]

Hello,

nextafter(x, numeric_limits<double>::max())

nextafter was added to C in C99, and is in C++'s TR1. It's also been
voted into the next version of the standard.

If your implementation doesn't supply it you'll probably have to do some
bit twiddling.

A portable /nextafter(x,numeric_limits<double>::max())/ could look like
the one given after the signature. Without any explicit or implied
warranty ;-)

Cheers,
Loic.
--
#include <float.h>

/*
* return the smallest double y, such that x < y
*/
double
my_nextafter(double x)
{
double y;
if (x>=0) {
y = x*(1+DBL_EPSILON);
} else {
y = x*(1-DBL_EPSILON);
}
double middle;
while (x < y ) {
middle = (x+y)/2;
if (x < middle && y!=middle) {
y = middle;
} else {
break;
}
}
return y;
}