0
\$\begingroup\$

I am doing a code for radix-4 booth encoding for 8*8 multiplication. The logic is correct and there are no errors or warning. The output am getting is totally unrelated. i have posted the code below

module comp8mul (a, b, outc, clk);
  input [7:0] a, b;
  input clk;
  output [4:0] outc;
  wire [8:0] a1;
  reg [8:0] a12;
  wire [11:0] b1;
  integer i;
  reg [2:0] c1;
  reg [11:0] ou1;
  reg [4:0] b11 [0:11];
  reg [4:0] s1;

  assign a1 = {1'b0,a};
  assign b1 = {1'b0,b,1'b0,1'b0};
  always @(posedge clk) begin
    for (i=11; i>=2'd2; i=i-(2'd2)) begin:rloop
      c1={b1[i-2'd2],b1[i-1'b1],b1[i]};

      case(c1)
      3'b000: begin
        ou1=11'b0;
        s1[(i-(2'd3))/2'd2]=1'b0;
      end
      3'b001: begin
        ou1=a1;
        s1[(i-(2'd3))/2'd2]=1'b0;
      end
      3'b010: begin
        ou1=a1;
        s1[(i-2'd3)/2'd2]=1'b0;
      end
      3'b011: begin
        a12=a1<<1;
        ou1=(~a12)+1'b1;
        s1[(i-(2'd3))/2'd2]=1'b1;
      end
      3'b100: begin
        a12=a1<<1'b1;
        ou1=(~a12)+1'b1;
        s1[(i-(2'd3))/2'd2]=1'b1;
      end
      3'b101: begin
        ou1=(~a1);
        s1[(i-(2'd3))/2'd2]=1'b1;
      end
      3'b110: begin
        ou1=(~a1);
        s1[(i-(2'd3))/2'd2]=1'b1;
      end
      3'b111: begin
        ou1=12'b0;
        s1[(i-(2'd3))/2'd2]=1'b0;
      end
      endcase
      if (((i-(2'd3))/2'd2)==3'd4) begin
        b11[((i-(2'd3))/(2'd2))]={(~s1[(i-(2'd3))/(2'd2)]),s1[(i-(2'd3))/(2'd2)],s1[(i-(2'd3))/2'd2],ou1};

      end
      if (((i-(2'd3))/(2'd2))==1'b1) begin
        b11[(i-(2'd3))/(2'd2)]={(~s1[(i-(2'd3))/(2'd2)]),ou1};
      end
      if(((i-(2'd3))/(2'd2))==1'b0) begin
        b11[((i-(2'd3))/(2'd2))]=ou1;
      end else
        b11[(i-(2'd3))/(2'd2)]={1'b1,(~s1[(i-(2'd3))/(2'd2)]),ou1};
    end
  end
  assign outc = b11[0];
endmodule

I should get five 12-bit partial products as outputs. I am assigning 1 row of the 2-d array to output and checking it.

\$\endgroup\$
  • \$\begingroup\$ Please format the code for clarity. \$\endgroup\$ – Arun M Apr 17 '13 at 12:00
  • 1
    \$\begingroup\$ @twinkle: I can not figure out what you are asking. What is the output you are getting that seems unexpected to you? \$\endgroup\$ – Wandering Logic Apr 17 '13 at 14:31
  • \$\begingroup\$ Show your testbench - what goes in, what comes out. What did you expect to get out? \$\endgroup\$ – Martin Thompson Apr 17 '13 at 16:01
  • 2
    \$\begingroup\$ Yes, you can use a case statement inside a for loop. However you do realize that the entire for loop will execute instantaneously (all 12 loops on the first posedge clock), right? \$\endgroup\$ – Tim Apr 17 '13 at 16:29
1
\$\begingroup\$

If "five 12-bit partial products as outputs" desired, then reg [4:0] b11 [0:11]; should be reg [11:0] b11 [0:4];.

Your b11 calculation is likely incorrect. There are missing elses before many of the if statements. Therefore, all except the final if condition with fall into the final else condition. You need to nest the else-ifs. Alternatively ,you can use another case statement. Also you need to control the ou1 range otherwise the upper assigned bits will be out of range for b11 and masked out.

Suggestion: (i-2'd3)/2'd2 and format variations of the same equations is used many places. Assign it to a variable, such as reg [2:0] c2, and call the variable (c2) instead of the equation. c2 should be assigned just like your c1. This will help readability of the code and lower chances of a typo.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.