Adding binary number literals in Verilog not giving expected results

Question

I am having some trouble displaying the result of my 8 bit adder verilog

module Adder(a,b,cin,s,co);
input [7:0]a;
input [7:0]b;
output [7:0]s;
output co;
wire [6:0] u;
input cin;

Carry c1(a[0],b[0],cin,s[0],u[0]);
Carry c2(a[1],b[1],u[0],s[1],u[1]);
Carry c3(a[2],b[2],u[1],s[2],u[2]);
Carry c4(a[3],b[3],u[2],s[3],u[3]);
Carry c5(a[4],b[4],u[3],s[4],u[4]);
Carry c6(a[5],b[5],u[4],s[5],u[5]);
Carry c7(a[6],b[6],u[5],s[6],u[6]);
Carry c8(a[7],b[7],u[6],s[7],co);

endmodule

module Carry(a,b,cin,s,co);

input wire a;
input wire b;
input wire cin;
output wire co;
output wire s;
assign co = (a & b) | (b & cin) | (a & cin);
assign s = (~a & ~b & cin) | (~a & b & ~cin) | (a & ~b & ~cin)| (a & b & cin);

endmodule

module testbench;

          reg [7:0]a;
          reg [7:0]b;
          reg cin;
          wire [7:0]s;
          wire co;
          Adder add(a, b, cin, s, co);
          initial begin
    $dumpfile("result.vcd");
    $dumpvars;
         a <= 00000010; b <= 00000010; cin <= 0;
     #5 

      $monitor("time=%4d: %b + %b + %b: sum = %b, carry = %b\n",$time,a,b,cin,s,co);
end
endmodule

although it adds everything correctly, it's not adding the numbers I originally wanted.

time=   5: 00001010 + 00001010 + 0: sum = 00010100, carry = 0

How can i fix it so that instead of adding those numbers, it would add the numbers i'd want. (a = 00000010; b = 00000010; cin = 0;) I already tried changing the numbers around and it does not work except when they're 00000001.

toolic · Accepted Answer · 2020-06-18 12:39:50Z

1

By default, Verilog interprets a numerical literal value as decimal. The value 00000010 is decimal 10. The $monitor statement uses %b, and it correctly displays the decimal value 10 as 1010. The sum of 10 + 10 is 20 (decimal), which is correctly displayed as 10100 (binary).

For Verilog to interpret 00000010 as binary, you need to specify the base as 'b00000010:

            a <= 'b00000010; b <= 'b00000010; cin <= 0;

Refer to IEEE Std 1800-2012, section 5.7.1 Integer literal constants.

The reason that 1 + 1 works is that 1 is the special case where 00000001 (decimal) is the same value as 'b00000001 (binary).

edited Jun 18, 2020 at 12:39

answered May 22, 2018 at 18:15

toolic

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