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32-bit adder question
von DSP_Arch_Student (Guest)

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Hello,

I am trying to write a code for a 32-bit adder. I am having trouble 
where the carry in signal is not be registered at all. Here is my code, 
can you guys give me some insight as what is wrong, I think I might not 
be understanding some fundamental concepts associated with VHDL. I am 
new to VHDL, so I welcome all sort of critiques.

Thanks,
D_A_S

Code:
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LIBRARY ieee;

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use ieee.std_logic_1164.all;

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ENTITY hmk2_32bitAdder IS

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  PORT(a, b : IN std_logic_vector(31 downto 0);

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     c : IN std_logic;

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     result : OUT std_logic_vector(31 downto 0);

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     cout : OUT std_logic);

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END ENTITY hmk2_32bitAdder;

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ARCHITECTURE adder_32bit OF hmk2_32bitAdder IS

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  SIGNAL c_internal : std_logic;

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  BEGIN

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    P1 : PROCESS (a, b, c, c_internal)

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     BEGIN

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       c_internal <= c;

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      for i in 0 to 31 loop

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        result(i) <= a(i) xor b(i) xor c_internal;

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          c_internal <= (a(i) and b(i)) or (a(i) and c_internal) or (b(i) and c_internal);

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      end loop;

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      cout <= c_internal;

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   END PROCESS P1;

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END ARCHITECTURE adder_32bit;


  


  




  

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    Re: 32-bit adder question
  


  

    

      von
      
        Lothar M.
            (Company: Titel)
        (lkmiller)
        (Moderator)
      
      


      
    


    

      
    

    

    

  
  


  

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DSP_Arch_Student wrote:
> where the carry in signal is not be registered at all.
An adder is a completly registerless, combinatorial design. So is yours: 
no clock, no registers.

And now: what do you mean with "not registered"?

One major flaw in your design ist, that you don't know about how signals 
behave in a process. They keep(!!) their value throughout(!!) the whole 
process. And at the end of the process they get the last assigned value. 
Think about that.

Lets start here with c_internal='0' and c='1'. On the right side of all 
the assginments througout the process c_internal is '0':
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    P1 : PROCESS (a, b, c, c_internal)

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     BEGIN

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       c_internal <= c; -- c_internal gets a new value that will possibly be transferred at the end of the process              

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      for i in 0 to 31 loop

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        result(i) <= a(i) xor b(i) xor c_internal; -- c_internal IS '0' HERE

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          c_internal <= (a(i) and b(i)) or (a(i) and c_internal) or (b(i) and c_internal); -- c_internal IS '0' HERE, it may get a new value at the end of the process

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      end loop;

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      cout <= c_internal; -- c_internal IS '0' here, so cout is '0' also

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   END PROCESS P1; -- And finally c_internal here gets the last assigned value from three lines above


In this case here c_internal must be a variable. Variables virtually 
take over the assigned value "immediately". Lets start again with 
c_internal='0' and c='1':
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ARCHITECTURE adder_32bit OF hmk2_32bitAdder IS

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  BEGIN

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    P1 : PROCESS (a, b, c, c_internal)

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     variable c_internal : std_logic;

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     BEGIN

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       c_internal := c; -- always initialise the variable!

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      for i in 0 to 31 loop -- surprise, surprise: now HERE c_internal IS '1'

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        result(i) <= a(i) xor b(i) xor c_internal;

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          c_internal := (a(i) and b(i)) or (a(i) and c_internal) or (b(i) and c_internal);

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      end loop;

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      cout <= c_internal;

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   END PROCESS P1;

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END ARCHITECTURE adder_32bit;

But be aware that variables in processes are not like variables in C 
programs (as VHDL is not a programming language, but a description 
language. just keep it in mind, you will find out further on)...

  


  




  

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    Re: 32-bit adder question
  


  

    

      von
      
        Sym (Guest)
      
      


      
    


    

      
    

    

    

  
  


  

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Apart from this 'educational' implementation, one would normally use 
ieee.numeric_std, and the '+' operator. Simply because it is much more 
readable, and - this is the main reason - the compiler can map it to 
arithmetic hardware which is faster and smaller.

  


  




  

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    Re: 32-bit adder question
  


  

    

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        Sigi (Guest)
      
      


      
    


    

      
    

    

    

  
  


  

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DSP_Arch_Student wrote:
> the carry in signal is not be registered at all

Registering all carries (+clock) would imply a
pipeline design, but I guess that's not
your aim.

The problem is, that you only use ONE!
signal for all carries, but incl. input Carry
you have 32 carries. Declare a new vector cy
and assign "cy[0] <= c" and in each stage
"cy[i+1] <= (a(i) and b(i)) or (a(i) and cy[i]) or (b(i) and cy[i]);

  


  




  

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    Re: 32-bit adder question
  


  

    

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        Sigi (Guest)
      
      


      
    


    

      
    

    

    

  
  


  

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sorry mistake, incl cout you need 33 carries:
cout <= cy[32]

  


  




  

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        DSP_Arch_Student (Guest)
      
      


      
    


    

      
    

    

    

  
  


  

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@Sigi, @lkmiller,

I know this is a little late on the reply, but thanks you guys for the 
help! I know what I did wrong now, a semester after the class is over! 
:D

As long as I know what I did wrong, I'm happy. So, thanks for the help 
again!!!

  


  




  

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