Right arithmetic shift does not work properly

Question

if(Op_ex == 5'b11011 || Op_ex==5'b11100 || Op_ex==5'b11101 || Op_ex==5'b11110 || Op_ex==5'b11111 )
 begin
    temp_carry = 1'b0;
    ans_ex = A   >>>   B;
    flag[3] = ans_ex[0]^ans_ex[1]^ans_ex[2]^ans_ex[3]^ans_ex[4]^ans_ex[5]^ans_ex[6]^ans_ex[7];  //flag[3] = parity
    flag[2] = 0;                                                  //flag[2] = overflow  
    flag[0] = 0;
    if(ans_ex == 8'b00000000)
        begin
            flag[1] = 1'b1;                                   //flag[1] = zero
        end
    else
        begin
            flag[1] = 1'b0;
        end
        data_out = reg_data;
 end         
end

If I put A = 8'hC0 and B = 8'h01 then it will give me output as 60, instead of e0 in hexadecimal .

A and B are 8 bit long and A = 8'hC0 and B=8'h01 so output would be 8'he0 but it gives me output as 8'h60 it only shifts registers to right not doing arithmetic operation — Jay Dangar, Sep 10 '15 at 18:34
You need to show the declarations of all signals in your example. An I think you meant to say that A and B are declared as `input [7:0] A,B;` there is a big difference. Please update your original post instead of adding comments. — dave_59, Sep 10 '15 at 18:49

score 1 · Accepted Answer · answered Sep 10 '15 at 19:44

1

A is unsigned. The sign extinction of an unsigned is always 0. Make A (possibly B well) a signed value. Two was to do so:

input signed [7:0] A,B
ans_ex = $signed(A) >>> B;

_{FYI:

(Op_ex == 5'b11011 || Op_ex==5'b11100 || Op_ex==5'b11101 || Op_ex==5'b11110 || Op_ex==5'b11111 ) can be simplified to (Op_ex >= 5'b11011)

flag[3] = ans_ex[0]^ans_ex[1]^ans_ex[2]^ans_ex[3]^ans_ex[4]^ans_ex[5]^ans_ex[6]^ans_ex[7]; can be simplified to: flag[3] = ^ans_ex; or flag[3] = ^ans_ex[7:0];}

answered Sep 10 '15 at 19:44

Greg

18,111
5
46
68

I try this but my output remains same it didi not give right arithmetic shift it gives only right shift not right arithmetci shift – Jay Dangar Sep 11 '15 at 03:35
thx greg it works on signed numbers but did not work on unsigned numbers – Jay Dangar Sep 11 '15 at 03:36

Right arithmetic shift does not work properly

1 Answers1