For finite values v0, v1 and value r in [0, 1] range, will the value v, computed as below, always belong to [v0, v1] range, or can it be (slightly) outside due to round off errors?
double v0; // Finite
double v1; // Finite
double r; // In [0, 1]
double v = v0 * r + v1 * (1.0 - r);
if (v0 <= v1)
assert(v0 <= v && v <= v1);
else
assert(v1 <= v && v <= v0);
Yes, it can be. Here's an example:
#include <assert.h>
int main() {
double v0 = 2.670088631008241e-307;
double v1 = 2.6700889402193536e-307;
double r = 0.9999999999232185;
double v = v0 * r + v1 * (1.0 - r);
if (v0 <= v1)
assert(v0 <= v && v <= v1);
else
assert(v1 <= v && v <= v0);
return 0;
}
This produces:
Assertion failed: (v0 <= v && v <= v1), function main, file b.cpp, line 12.
The value of v computed in this case is:
2.67009e-307