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treitmey
Joined: 23 Jan 2004
Posts: 1094
Location: Appleton,WI USA
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| SQRT of an 32bit integer |
 Posted: Tue Nov 30, 2004 9:14 am |
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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);
}
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Last edited by treitmey on Wed Dec 24, 2008 8:45 am; edited 2 times in total |
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mvaraujo
Joined: 20 Feb 2004
Posts: 59
Location: Brazil
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| Posted: Tue Nov 30, 2004 12:36 pm |
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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 |
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PCM programmer
Joined: 06 Sep 2003
Posts: 21708
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| Posted: Tue Nov 30, 2004 1:23 pm |
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But you need keywords so people can search for your post.
I'll put them in:
square root
squareroot |
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kender
Joined: 09 Aug 2004
Posts: 768
Location: Silicon Valley
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| Posted: Tue Dec 23, 2008 9:48 pm |
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| 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
_________________
Read the label, before opening a can of worms. |
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OzanGazi
Joined: 03 Mar 2008
Posts: 6
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| Posted: Fri Jan 02, 2009 7:44 am |
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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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