# integer 32bit/32bit divide code

Started by January 5, 2006
```Hi users,
I am looking for 32bit/32bit divide code.
Now, I am implementing the PNS random generator for MPEG4 AAC.
I find some divide floating point code in the PNS as followings:
scale = 0x10000000/energy
Of course, energy represents the 32bit fixed point formant. I think two
variables regard as a Q0 format.
Does anybody give me the divide code or explain the theory of integer
divide?

I appreciate it.

```
```My dear tommy,

The theory of the integer division is explained in the 3rd class of the
elementary school.

I feel real sorry about the software engineers like you...

VLV

tommy wrote:
> Hi users,
> I am looking for 32bit/32bit divide code.
> Now, I am implementing the PNS random generator for MPEG4 AAC.
> I find some divide floating point code in the PNS as followings:
>                 scale = 0x10000000/energy
> Of course, energy represents the 32bit fixed point formant. I think two
> variables regard as a Q0 format.
> Does anybody give me the divide code or explain the theory of integer
> divide?
>
> I appreciate it.
>
```
```tommy wrote:
> Hi users,
> I am looking for 32bit/32bit divide code.
> Now, I am implementing the PNS random generator for MPEG4 AAC.
> I find some divide floating point code in the PNS as followings:
>                 scale = 0x10000000/energy
> Of course, energy represents the 32bit fixed point formant. I think two
> variables regard as a Q0 format.
> Does anybody give me the divide code or explain the theory of integer
> divide?
>
> I appreciate it.

There are many ways to do division on a computer. The best way depends
on the hardware that your computer has. If it has a good multiplier, one
good way would use Newton's method to find the reciprocal, then multiply.

You need to understand your resources in order to use them well. Are you
aware that the width of your ALU matters? You haven't told it to us.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```The post by the Troll does has an element of insight to it; the
reference to 3rd year elementary school.  By this I mean, this is about
the time that students are introduced to long division.

Integer division can be worked out in the same fashion.  As an example,
lets divide 100b by 10b (or 8 by 4 in decimal).  Following the
(elementary school) algorithm, we would first ask, how many times does
"10" go into "10", with the answer being 1, which is the first digit of
the answer.  We would then put '1' times 10 below the 100b and subtract
leaving a (remainder) of '0' as a  carry down.  Next we realize that
"10" will go into "0", "0" times, so the next digit is a 0.  We could
either stop here, or place a decimal (binary) point and continue to
append zeros and continue the process.   At the end we would wind up
the the binary result of 10.0  , where the '.' is the binary point.  In
decimal 10b is '2' which is the correct answer.

I have seen software algorithms implemented in this fashion, though I
am sure that a search would turn up shortcuts that are more efficient.

As a user / designer, it is important to understand that the computer
will simply perform the raw calculation on the numbers and it is your
responsibility to make sure that the numbers make sense.  One potential
problem is that a 32 bit number divided by a 32 bit number could very
well "overflow".  For example, in unsigned 2s complement for
simplicity, lets say we have the maximum value and that it represents
as close to 1 as we can get (.999999etc) = 0x(.)FFFFFFFF.  If we divide
this by .5 = 0x(.)10000000 where I have included the fictational binary
points for clarity, we are i n trouble.  The answer, which would be 2
wont fit in our result.  Consequently, one normally divides a 2N bit
number by a N bit number to ensure that there is enough space for the
result.

```
```Ooops, I noticed an error in my above post, 100b and 10b are 4 and 2,
not 8 and 4.  Initially I was going to use 1000 and 100 but decided to
simplify the example.

```
```See rules for fixed-point division in

http://www.digitalsignallabs.com/fp.pdf

--RY

#include <stdint.h>

uint64_t Divide32(uint32_t y, uint32_t x)
{
uint16_t n;
uint64_t remainder;
uint64_t divisor;

remainder = x;
divisor = (uint64_t)y << 32;

for (n = 0; n < 32; n++)
{
divisor = divisor >> 1;
if (remainder >= divisor)
{
remainder -= divisor;
answer |= (uint64_t)1 << (63 - n);
}
}

for (n = 0; n < 32; n++)
{
remainder = remainder << 1;
if (remainder >= divisor)
{
remainder -= divisor;
answer |= (uint64_t)1 << (31 - n);
}
}

```
```"tommy" <parcor@gmail.com> writes:

> Hi users,
> I am looking for 32bit/32bit divide code.
> Now, I am implementing the PNS random generator for MPEG4 AAC.
> I find some divide floating point code in the PNS as followings:
>                 scale = 0x10000000/energy
> Of course, energy represents the 32bit fixed point formant. I think two
> variables regard as a Q0 format.
> Does anybody give me the divide code or explain the theory of integer
> divide?

I describe how to perform fixed-point division here:

http://www.digitalsignallabs.com/fp.pdf

and the routine you seek (for unsigned ints) is here:

uint64_t Divide32(uint32_t y, uint32_t x)
{
uint16_t n;
uint64_t remainder;
uint64_t divisor;

remainder = x;
divisor = (uint64_t)y << 32;

for (n = 0; n < 32; n++)
{
divisor = divisor >> 1;
if (remainder >= divisor)
{
remainder -= divisor;
answer |= (uint64_t)1 << (63 - n);
}
}

for (n = 0; n < 32; n++)
{
remainder = remainder << 1;
if (remainder >= divisor)
{
remainder -= divisor;
answer |= (uint64_t)1 << (31 - n);
}
}

}
--
%  Randy Yates                  % "She's sweet on Wagner-I think she'd die for Beethoven.
%% Fuquay-Varina, NC            %  She love the way Puccini lays down a tune, and
%%% 919-577-9882                %  Verdi's always creepin' from her room."
%%%% <yates@ieee.org>           % "Rockaria", *A New World Record*, ELO
```
```Randy Yates wrote:
> See rules for fixed-point division in
>
>   http://www.digitalsignallabs.com/fp.pdf
>
> --RY
>
> #include <stdint.h>
>
> uint64_t Divide32(uint32_t y, uint32_t x)
> {
>    uint16_t n;
>    uint64_t remainder;
>    uint64_t divisor;
>
>    remainder = x;
>    divisor = (uint64_t)y << 32;
>
>    for (n = 0; n < 32; n++)
>    {
>       divisor = divisor >> 1;
>       if (remainder >= divisor)
>       {
>          remainder -= divisor;
>          answer |= (uint64_t)1 << (63 - n);
>       }
>    }
>
>    for (n = 0; n < 32; n++)
>    {
>       remainder = remainder << 1;
>       if (remainder >= divisor)
>       {
>          remainder -= divisor;
>          answer |= (uint64_t)1 << (31 - n);
>       }
>    }
>

I used to do this in assembly, shifting the quotient into a register a
bit at a time as the dividend shifted out of the same register. I could
do this on most DSPs, too, but their multipliers make finding the
reciprocal and multiplying much more efficient. Repeating from an
earlier post on mine in the thread, "You need to understand your
resources in order to use them well."

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```Jerry Avins <jya@ieee.org> writes:
> [...]
> Repeating from an earlier post on mine in the thread, "You need to
> understand your resources in order to use them well."

Yes, that's true. However...

I don't view the OP as being unclear in his original request. He
asked for a 32-bit division and I gave him one.

I have been frustrated in the past by folks attempting to
"second-guess" me and I am treating tommy the way I would
like to be treated: to simply answer the question that was
--
%  Randy Yates                  % "Remember the good old 1980's, when
%% Fuquay-Varina, NC            %  things were so uncomplicated?"
%%% 919-577-9882                % 'Ticket To The Moon'
%%%% <yates@ieee.org>           % *Time*, Electric Light Orchestra
```
```Randy Yates wrote:
> Jerry Avins <jya@ieee.org> writes:
>
>>[...]
>>Repeating from an earlier post on mine in the thread, "You need to
>>understand your resources in order to use them well."
>
>
> Yes, that's true. However...
>
> I don't view the OP as being unclear in his original request. He
> asked for a 32-bit division and I gave him one.
>
> I have been frustrated in the past by folks attempting to
> "second-guess" me and I am treating tommy the way I would
> like to be treated: to simply answer the question that was

You gave him good cross-platform code, and if he hasn't yet thanked you
for it, I'm sure he will. His application might not tolerate the time it
takes, and I want him to know that there may well be more efficient ways
to divide on his computer, and maybe even an algorithm (fraction
saving?) that avoids the need.

If you took my comment as a slur on you, I framed it badly.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```