(no title)
ambrop7 | 6 years ago
Be aware that +0.0 and -0.0 are different floating point values but represent the same real number, so +0.0 == -0.0 follows.
People who say == means nothing for floating point and you always need epsilon checks are wrong, plain and simple. == is very well defined. Don't confuse the definition of floating point operations with common practices for using them effectively.
You can iterate through all non-NaN values and check that successive ones are indeed not equal:
#include <math.h>
#include <stdint.h>
#include <assert.h>
#include <stdio.h>
#include <inttypes.h>
int main()
{
float x = (float)-INFINITY;
uint64_t count = 1;
while (x != (float)INFINITY) {
float y = nextafterf(x, (float)INFINITY);
assert(y != x);
x = y;
++count;
}
printf("Found %" PRIu64 " floats.\n", count);
return 0;
}
$ gcc -std=c99 -O3 a.c -lm -o a
$ ./a
Found 4278190081 floats.
(a little bit harder for doubles)Interestingly, this only finds one zero (-0.0), hence the assert doesn't actually fail around zero.
thanatropism|6 years ago
(a) get x casted to a float
(b) get `div` instead (like in Python 2) and obtain a false result
(c) get cursed out with a type error.
These three options are available because it's possible to determine whether a given real number is representable as an int or not. This is not possible with floats.
ambrop7|6 years ago
What you can do is:
- determine if a float is an integer (trunc(x) == x),
- convert a float to a certain integer type with some kind of rounding, or get an error if it's out of range (see my comment with double_to_uint64),
- convert a float to a certain integer type exactly, or get an error if it's not representable (e.g. by doing both of the above).
The basic reason that so many people fail to use floats correctly is that they act like operations on floats are equivalent to operations on the real numbers they represent, when in fact they are usually defined as the operation on real numbers rounded to a representable value.