Fast Inverse Square Root on x64
Here is an implementation for double precision floats:
#include <cstdint>
double invsqrtQuake( double number )
{
double y = number;
double x2 = y * 0.5;
std::int64_t i = *(std::int64_t *) &y;
// The magic number is for doubles is from https://cs.uwaterloo.ca/~m32rober/rsqrt.pdf
i = 0x5fe6eb50c7b537a9 - (i >> 1);
y = *(double *) &i;
y = y * (1.5 - (x2 * y * y)); // 1st iteration
// y = y * ( 1.5 - ( x2 * y * y ) ); // 2nd iteration, this can be removed
return y;
}
I did a few tests and it seems to work fine
Originally Fast Inverse Square Root was written for a 32-bit float, so as long as you operate on IEEE-754 floating point representation, there is no way x64 architecture will affect the result.
Note that for "double" precision floating point (64-bit) you should use another constant:
...the "magic number" for 64 bit IEEE754 size type double ... was shown to be exactly 0x5fe6eb50c7b537a9
Yes, it works if using the correct magic number and corresponding integer type. In addition to the answers above, here's a C++11 implementation that works for both double
and float
. Conditionals should optimise out at compile time.
template <typename T, char iterations = 2> inline T inv_sqrt(T x) {
static_assert(std::is_floating_point<T>::value, "T must be floating point");
static_assert(iterations == 1 or iterations == 2, "itarations must equal 1 or 2");
typedef typename std::conditional<sizeof(T) == 8, std::int64_t, std::int32_t>::type Tint;
T y = x;
T x2 = y * 0.5;
Tint i = *(Tint *)&y;
i = (sizeof(T) == 8 ? 0x5fe6eb50c7b537a9 : 0x5f3759df) - (i >> 1);
y = *(T *)&i;
y = y * (1.5 - (x2 * y * y));
if (iterations == 2)
y = y * (1.5 - (x2 * y * y));
return y;
}
As for testing, I use the following doctest in my project:
#ifdef DOCTEST_LIBRARY_INCLUDED
TEST_CASE_TEMPLATE("inv_sqrt", T, double, float) {
std::vector<T> vals = {0.23, 3.3, 10.2, 100.45, 512.06};
for (auto x : vals)
CHECK(inv_sqrt<T>(x) == doctest::Approx(1.0 / std::sqrt(x)));
}
#endif