Computing square root from fixed point

Question

I have been trying to compute the square root from a fixed point data-type <24,8>.
Unfortunately nothing seems to work.
Does anyone know how to do this fast and efficient in C(++)?

Answer 1

Here is a prototype in python showing how to do a square root in fixed point using Newton's method.

import math

def sqrt(n, shift=8):
     """
     Return the square root of n as a fixed point number.  It uses a
     second order Newton-Raphson convergence.  This doubles the number
     of significant figures on each iteration.

     Shift is the number of bits in the fractional part of the fixed
     point number.
     """
     # Initial guess - could do better than this
     x = 1 << shift // 32 bit type
     n_one = n << shift // 64 bit type
     while 1:
         x_old = x
         x = (x + n_one // x) // 2
         if x == x_old:
             break
     return x

def main():
    a = 4.567
    print "Should be", math.sqrt(a)
    fp_a = int(a * 256)
    print "With fixed point", sqrt(fp_a)/256.

if __name__ == "__main__":
    main()

When converting this to C++ be really careful about the types - in particular n_one needs to be a 64 bit type or otherwise it will overflow on the <<8 bit step. Note also that // is an integer divide in python.