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SQRT of an 32bit integer

 
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treitmey



Joined: 23 Jan 2004
Posts: 1094
Location: Appleton,WI USA

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SQRT of an 32bit integer
PostPosted: Tue Nov 30, 2004 9:14 am     Reply with quote

SQRT of an 32bit integer
square root sqrt
squareroot
shamlessly stolen from general forum. Looks like it belongs here.
Code:
long uisqrt32(int32 r)
{
   int32 t,b,c=0;

   for (b=0x10000000;b!=0;b>>=2) {
      t = c + b;
      c >>= 1;
      if (t <= r) {
         r -= t;
         c += b;
      }
   }
   return(c);
}



Last edited by treitmey on Wed Dec 24, 2008 8:45 am; edited 2 times in total
mvaraujo



Joined: 20 Feb 2004
Posts: 59
Location: Brazil

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PostPosted: Tue Nov 30, 2004 12:36 pm     Reply with quote

treitmey,

Did you have opportunity to check it out?

Sorry, I've posted there, but here is the place for it. When I logged in to post the code, you already did.

Marcus
PCM programmer



Joined: 06 Sep 2003
Posts: 20107

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PostPosted: Tue Nov 30, 2004 1:23 pm     Reply with quote

But you need keywords so people can search for your post.
I'll put them in:

square root

squareroot
kender



Joined: 09 Aug 2004
Posts: 768
Location: Silicon Valley

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PostPosted: Tue Dec 23, 2008 9:48 pm     Reply with quote

mvaraujo wrote:
Did you have opportunity to check it out?


I've just black-box tested it. It seems to be working. The error is on the low side.



Another keyword: sqrt
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OzanGazi



Joined: 03 Mar 2008
Posts: 6

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PostPosted: Fri Jan 02, 2009 7:44 am     Reply with quote

i have just tested the code and its working for 30bit integer max.

Function returns 32767 while r > 30bit..

But why??
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