#include
#define BITSPERLONG 32
#define TOP2BITS(x) ((x & (3 << (BITSPERLONG-2))) >> (BITSPERLONG-2))
/* usqrt:
ENTRY x: unsigned long
EXIT returns floor(sqrt(x) * pow(2, BITSPERLONG/2))
Since the square root never uses more than half the bits
of the input, we use the other half of the bits to contain
extra bits of precision after the binary point.
EXAMPLE
suppose BITSPERLONG = 32
then usqrt(144) = 786432 = 12 * 65536
usqrt(32) = 370727 = 5.66 * 65536
NOTES
(1) change BITSPERLONG to BITSPERLONG/2 if you do not want
the answer scaled. Indeed, if you want n bits of
precision after the binary point, use BITSPERLONG/2+n.
The code assumes that BITSPERLONG is even.
(2) This is really better off being written in assembly.
The line marked below is really a "arithmetic shift left"
on the double-long value with r in the upper half
and x in the lower half. This operation is typically
expressible in only one or two assembly instructions.
(3) Unrolling this loop is probably not a bad idea.
ALGORITHM
The calculations are the base-two analogue of the square
root algorithm we all learned in grammar school. Since we're
in base 2, there is only one nontrivial trial multiplier.
Notice that absolutely no multiplications or divisions are performed.
This means it'll be fast on a wide range of processors.
*/
struct int_sqrt {
unsigned sqrt,
frac;
};
void usqrt(unsigned long x, struct int_sqrt *q)
{
unsigned long a = 0L; /* accumulator */
unsigned long r = 0L; /* remainder */
unsigned long e = 0L; /* trial product */
int i;
for (i = 0; i < BITSPERLONG; i++) /* NOTE 1 */
{
r = (r << 2) + TOP2BITS(x); x <<= 2; /* NOTE 2 */
a <<= 1;
e = (a << 1) + 1;
if (r >= e)
{
r -= e;
a++;
}
}
memcpy(q, &a, sizeof(long));
}
#ifdef TEST
#include
#include
main(void)
{
int i;
unsigned long l = 0x3fed0169;
struct int_sqrt q;
for (i = 0; i < 101; ++i)
{
usqrt(i, &q);
printf("sqrt(%3d) = %2d, remainder = %2d\n",
i, q.sqrt, q. frac);
}
usqrt(l, &q);
printf("\nsqrt(%lX) = %X, remainder = %X\n", l, q.sqrt, q.frac);
return 0;
}
#endif /* TEST */