Xilinx does not support real math for synthesis. A short search with 
your prefered internet search engine should have shown that.

and it gave me the same error,but when i divide by 3
i hope some one help us

yes y r right ,but how i can solve it?

You may implement - or rather describe - a divider on your own. Possibly 
there is an description on the internet. Just give your search engine a 
try.
A good starting point is also the Hamburg VHDL Archive.

I forgot: There is a IP-Generator for dividers in Xilinx, too. You may 
use this either.

what is the datatype of norm_dis? If it would be integer, the solution 
would be quite simple: Just calculate 1000*sum and ignore the lower 15 
bits.

If it would be floating-point (i.e. real), then you have to accept that 
actual hardware/development-software of FPGAs is not powerfull enough, 
to implement this calculation right away. A good solution then might be 
to use fixpoint-calculation instead of floating point calculation.

I.e. just replace the expression
   1000*sum/32768 by
   1000*sum
and "keep in mind", that you have shifted the position of the 
"decimal-point" 15 positions to the left. (well: it's not a decimal 
point shift of 15 decimal positions, we just shift 15 dual positions as 
2^15=32768)

thanks a lot ,but i use 1000 because if i divide without the result 
would be float so i get ride of it by multiply by 1000

yes norm_dis is integer

meno wrote:
> thanks a lot ,but i use 1000 because if i divide without the result
> would be float so i get ride of it by multiply by 1000

Well, that's not exaclty true. The integer division of 255/32768 does 
not result in a floating point value. It results in the integer constant 
0. The integer calculation of 1000*sum/32768 results in an integer in 
the range 0..7 (cause 1000*255/32768 is 7 plus some remainder).

So if the integer result is what you are interested in: just take the 
multiplication of 1000*sum (which is synthesizable, either by logic 
elements of with a multiplier IP-core) and ignore the lowest 15 bits of 
the result. Ignoring 15 lsbs is equivalent to an integer division by 
32768.

Maybe you could simplify your calculation further: the multiplication 
with 1000 is very close to the multiplication with 1024. And the 
calculation
   1024*sum/32768
is nothing else but ignoring the lowest 5 bits of sum.

thanks a lot all