Verilog Construction of Two's Complement Comparator

Question

I am trying to write some simple verilog code for a comparator of two 4 bit two's complement numbers. I have two 4-bit inputs (A[3:0], B[3:0]), and 3 outputs (AeqB, AgtB, AltB) to show if A and B are equal, if A is greater than B, or A is less than B. There is also a third input named sign, which if 0 means that the numbers are unsigned, and if 1, the numbers are signed.

So i know that two signed two's complement numbers can be compared by subtracting them, but i cannot get that to work properly in my design. Here is what i have tried:

if(sign==0)
 begin

    if(({sign,A}-{sign,B})==0)
        AeqB = 1;
    else if(({sign,A}-{sign,B}) > 0)
        AgtB = 1;
    else if (({sign,A}-{sign,B}) < 0
        AltB = 1;
end

It seems as though this should work. I concatenate the sign bit to the front of the 4 bit numbers, subtract them, and then check to see if they are greater than or equal to zero. If A-B<0, then B is less than A because they are both negative numbers.

However when i simulate this design, it is correct whenever A=B, but shows AgtB in every other case, never AltB.

Any ideas on what i am doing wrong?

Answer 1

I am not sure what you are doing with {sign,A}. This would force the numbers to be negative when they have a signed format. Unless sign forces them to be negative?

you could define inputs to be signed or force a signed comparison, and 0 pad the msb in the unsigned case to share hardware, what you have implied is three subtractors in parallel. Synthesis might do a great job and share the hardware but you would have to check each time that it has given you what you want.

if (sign) begin
   A_i = {A[3], A};
   B_i = {B[3], B};
end
else begin
   A_i = {1'b0, A};
   B_i = {1'b0, B};
end

AgtB = $signed(A_i) > $signed(B_i) ;
AltB = ~AgtB ;

When adding subtracting numbers the result the bit growth is the largest bit width input +1. for unsigned 0 pad and interrupt result as unsigned. For signed numbers, sign extend by repeating the MSB.

To help with the understanding of 2's complement I have included some 4 bit examples of sign unsigned arithmetic.

Unsigned arithmetic:

   3 : (0)0011 
 + 1 : (0)0001
 = 4 :  0 0100    

Large Unsigned arithmetic:

  15 :  (0)1111 //Zero pad for correct bitwidths
 + 1 :  (0)0001
 =16 :   1 0000

Large Signed (Overflow):
On the result different (01) MSBs indicate overflow for truncation back down to 4 bits

   7 : (0)0111
  +1 : (0)0001
  =8 :  0 1000 bits

Subtraction:

   7 : (0)0111
  -1 : (1)1111 //(twos complement of 1)
  // Sum the bits as you did for unsigned
  =6 :  0 0110