This code takes advantage of IEEE format of floating point numbers to
get a close approximation. Then two iterations of Newton's method get
about 12 digits of accuracy.
Don
float sqrt(float z) {
union
{
int tmp;
float f;
} u;
u.f = z;
u.tmp -= 1<<23;
u.tmp >>= 1;
u.tmp += 1<<29;
u.f = u.f-(u.f*u.f-z)/(2.0*u.f);
return u.f-(u.f*u.f-z)/(2.0*u.f);
}
On Aug 30, 1:07 am, Raghavan <its...@gmail.com> wrote:
> how to design this logic effectively?
>
> double squareRoot(int num){
>
> }
>
> --
> Thanks and Regards,
> Raghavan KL
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