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);
CodePudding user response:
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
CodePudding user response:
Various comments and an answer provide examples that v
may fall outside the range.
A candidate fix is to round-trip the r
to make certain r
and 1.0 - r
add up to 1.0.
r = 1.0 - r;
r = 1.0 - r;
v = v0 * r v1 * (1.0 - r);
So far I have been unable to fail OP's range test.