I came across this piece of code. But I am not sure how is CRC calculated here. I know the theoretical way of calculating CRC but I guess here it is using some kind of different logic, maybe. Please correct me.
r_o[14:13] <= r_o[13:12];
r_o[12] <= r_o[11]^r_o[15]^x16;
r_o[11] <= r_o[10];
r_o[10:6] <= r_o[9:5];
r_o[5] <= r_o[4]^r_o[15]^x16;
r_o[4] <= r_o[3];
r_o[3:1] <= r_o[2:0];
r_o[0] <= r_o[15]^x16;
Well, if you know about the theoretical way to do we can start with
Assuming u is the input bit, and that every iteration you calculate r = r * x + u*x^16, remember that in the given ring x^16 = x^12 + x^5 + 1, a direct implementation of this would be
parameter X16_MASK = 16'h1021; // (1 << 12) ^ (1 << 5) ^ 1
function [15:0] step(input [15:0] prev_state, input u);
logic [15:0] state;
begin
state = {prev_state[14:0], 1'b0};
if(state[15]) state = state ^ X16_MASK;
if(u) state = state ^ X16_MASK;
step = state;
end
endfunction
This could simply be written as {r_o[14:0], 1'b0} ^ (X16_MASK * (r_o[15] ^ u)) and let the synthesis to optimize whatever is necessary, it should be able to simplify the multiplication by a 1-bit signal. Now check the positions where the mask has an effect you will get to assignments above.
I would simplify
r_o[11] <= r_o[10];
r_o[10:6] <= r_o[9:5];
to r_o[11:6] = r_o[10:5]
and
r_o[4] <= r_o[3];
r_o[3:1] <= r_o[2:0];
to r_o[4:1] = r_o[3:0]
In the code you presented I am missing the assignment to r_o[15].
So you could say
r_o[15:13] <= r_o[14:12];
r_o[12] <= r_o[11]^r_o[15]^x16;
r_o[11:6] <= r_o[10:5];
r_o[5] <= r_o[4]^r_o[15]^x16;
r_o[4:1] <= r_o[3:0];
r_o[0] <= r_o[15]^x16;
And if you like the one linear bit packing
r_o <= {r_o[14:12], r_o[11]^r_o[15]^x16, r_o[10:5], r_o[4]^r_o[15]^x16,r_o[3:0], r_o[15]^x16}
This question already has answers here:
What is the difference between reg and wire in a verilog module?
(3 answers)
Closed 2 years ago.
I have some code in VHDL I am trying to convert to Verilog.
The VHDL code works fine
library ieee;
use ieee.std_logic_1164.all;
entity find_errors is port(
a: bit_vector(0 to 3);
b: out std_logic_vector(3 downto 0);
c: in bit_vector(5 downto 0));
end find_errors;
architecture not_good of find_errors is
begin
my_label: process (a,c)
begin
if c = "111111" then
b <= To_StdLogicVector(a);
else
b <= "0101";
end if;
end process;
end not_good;
The Verilog code I have, gets errors "Illegal reference to net "bw"."
and
Register is illegal in left-hand side of continuous assignment
module find_errors(
input [3:0]a,
output [3:0]b,
input [5:0]c
);
wire [0:3]aw;
wire [3:0]bw;
reg [5:0]creg;
assign aw = a;
assign b = bw;
assign creg = c;
always #(a,c)
begin
if (creg == 4'b1111)
bw <= aw;
else
bw <= 4'b0101;
end
endmodule
It looks pretty close but there are a few things that are problems/errors that need to be fixed, see inline comments in fixed code:
module find_errors(
input wire [3:0] a, // Better to be explicit about the types even if
// its not strictly necessary
output reg [3:0] b, // As mentioned in the comments, the error youre
// seeing is because b is a net type by default; when
// describing logic in any always block, you need to
// use variable types, like reg or logic (for
// SystemVerilog); see the comment for a thread
// describing the difference
input wire [5:0] c);
// You dont really need any local signals as the logic is pretty simple
always #(*) begin // Always use either always #(*), assign or
// always_comb (if using SystemVerilog) for combinational logic
if (c == 6'b111111)
b = a; // For combinational logic, use blocking assignment ("=")
// instead of non-blocking assignment ("<="), NBA is used for
// registers/sequential logic
else
b = 4'b0101;
end
endmodule
aw and creg are unnecessary, and bw needs to be declared as reg.
module find_errors(
input [3:0] a,
output [3:0] b,
input [5:0] c
);
reg [3:0] bw;
assign b = bw;
always #(a,c)
begin
if (c == 4'b1111)
bw <= a;
else
bw <= 4'b0101;
end
endmodule
Since there is no sequential logic, you don't even need an always block:
module find_errors(
input [3:0] a,
output [3:0] b,
input [5:0] c
);
assign b = (c == 4'b1111) ? a : 4'b0101;
endmodule
I am trying to determine how to turn this code into a 4-bit adder/subtractor using a fulladder. Right now it is doing the adding but I don't know how to do the subtract part.
module Adder #(parameter N = 4)(
output wire [N-1:0] sum, // sum
output wire co, // carry
input wire [N-1:0] x,
input wire [N-1:0] y,
input wire is_sub;
);
wire [N:0] c;
assign c[0] = 1'b0;
assign co = c[N];
genvar i;
generate
for (i = 0; i < N; i=i+1)
begin : counter_gen_label
FA FAInst (
.s(sum[i]),
.co(c[i+1]),
.a(x[i]),
.b(y[i]),
.cin(c[i]),
.is_sub(is_sub)
);
end
endgenerate
endmodule
module FA(
output reg s,
output reg co,
input wire a,
input wire b,
input wire cin,
input wire is_sub
);
always #(*)
begin
s = a ^ b ^ cin;
co = (a & b) | (a & cin) | (b & cin);
end
endmodule
How would I go by doing the subtraction inside the FA module?
Thanks!
FA does not need to use is_sub input.
Replace c[0] = 1'b0; with c[0] = is_sub;, and .b(y[i]) with .b(y[i] ^ is_sub).
This is from x - y = x + y' + 1 where y' means inverted y.
Here's my verilog code about add and shift multiplying
when I compile and Initialze and adding the inputs and outputs to get waveforms and simulating them, I dont see any results, everything is z... what is the problem?
module multi(a, b, ans);
input [3:0] a;
input [3:0] b;
output reg [15:0] ans;
reg [15:0] aa;
reg [15:0] bb;
reg [15:0] tmp=0;
reg flag = 1'b1;
always #( a, b)
begin
aa = a;
bb = b;
while ( flag == 1'b1 )
begin
if( bb[0] == 1'b1 )
tmp = tmp + aa;
aa = aa << 1;
bb = bb >> 1;
if ( bb==0 )
flag = 1'b0;
end
ans = tmp;
end
endmodule
There are a number of things that look strange with this code.
First is that you have no clock input, but are attempting to do everything with combinatorial logic.
Second is that setting flag to 1 in the reg statement will mean that your module is only capable of doing a single multiplication. By the way, it is more normal (especially for ASIC design) to use a reset signal than use this initialisation in a reg line.
Third is that a 4 bit number times a 4 bit number will result in an 8bit answer, not 16bit.
In any case, unless you are working at very high speeds you should be able to perform a multiply in a single cycle.
Here are a couple of ways of writing this code more naturally:
Combinatorial Style
module multi(a, b, ans);
input [3:0] a;
input [3:0] b;
output reg [7:0] ans;
always #(*)
begin
ans = a * b;
end
endmodule
Clocked style
module multi(clk, a, b, ans);
input [3:0] a;
input [3:0] b;
output reg [7:0] ans;
always #(posedge clk)
begin
ans <= a * b;
end
endmodule
Your input is 4 bits wide and your assigning to a 16 bit variable, the top 12 bits are unassigned ie x.
input [3:0] a;
reg [15:0] aa;
//...
aa = a;
To assign all bits of aa try some thing like:
aa = {12'b0, a};
//{} is bit concatenation
Or to sign extend a to 16 bits, repeat the MSB 12 times using {width{value}} replication:
aa = {{12{a[3]}}, a};
I'm studying Verilog and here's my first ALU.
I can't understand why the output does not display in the tester block.
Sample outputs(scroll horizontally):
FAIL: a=00010010000101010011010100100100, b=11000000100010010101111010000001, op=101, z=zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz, expect=11010010100111010111111110100101
FAIL: a=10000100100001001101011000001001, b=10110001111100000101011001100011, op=101, z=zzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz, expect=10110101111101001101011001101011
Why isn't z calculated?
ALU
module yAlu(z, ex, a, b, op);
input [31:0] a, b;
input [2:0] op;
output[31:0] z;
output ex;
wire [31:0] andRes, orRes, arithmRes, slt;
wire cout;
assign slt = 0; // not supported
assign ex = 0; // not supported
and myand[31:0] (andRes, a, b);
or myor[31:0](orRes, a, b);
//Instantiating yArith adder/subtractor from addSub.v
yArith addSub(arithmRes, cout, a, b, op[2]);
//Instantiating 4-to-1 32-bit multiplexor from 4to1Mux.v
yMux4to1 multiplexor(z, andRes, orRes, arithmRes, slt, op[1:0]);
endmodule
MULTIPLEXORS:
// 1-bit 2 to 1 selector
module yMux1(z, a, b, c);
output z;
input a, b, c;
wire notC, upper, lower;
not my_not(notC, c);
and upperAnd(upper, a, notC);
and lowerAnd(lower, c, b);
or my_or(z, upper, lower);
endmodule
//--------------------------------------------
// n-bit 2 to 1 selector
module yMux(z, a, b, c);
parameter SIZE = 2;
output [SIZE-1:0] z;
input [SIZE-1:0] a, b;
input c;
yMux1 mine[SIZE-1:0] (z, a, b, c);
endmodule
//--------------------------------------------
// n-bit 4-to-1 multiplexor
module yMux4to1(z, a0, a1, a2, a3, c);
parameter SIZE = 32;
output [SIZE-1:0] z;
input [SIZE-1:0] a0, a1, a2, a3;
input [1:0] c;
wire [SIZE-1:0] zLo, zHi;
yMux #(.SIZE(32)) lo(zLo, a0, a1, c[0]);
yMux #(.SIZE(32)) hi(zLo, a2, a3, c[0]);
yMux #(.SIZE(32)) final(zLo, zLo, zHi, c[1]);
// c in array is important (see LabL4.v page)
endmodule
//----------------------------------------------
ADDER/SUBTRACTOR BLOCK:
// A simple 1-bit full adder
module yAdder1(z, cout, a, b, cin);
output z, cout;
input a, b, cin;
xor left_xor(tmp, a, b);
xor right_xor(z, cin, tmp);
and left_and(outL, a, b);
and right_and(outR, tmp, cin);
or my_or(cout, outR, outL);
endmodule
//----------------------------------------------
// 32-bit adder with 1 bit carry
module yAdder(z, cout, a, b, cin);
output [31:0] z;
output cout;
input [31:0] a, b;
input cin;
wire [31:0] in, out;
yAdder1 adder[31:0](z, out, a, b, in);
assign in[0] = cin;
assign in[1] = out[0];
assign in[2] = out[1];
assign in[3] = out[2];
assign in[4] = out[3];
assign in[5] = out[4];
assign in[6] = out[5];
assign in[7] = out[6];
assign in[8] = out[7];
assign in[9] = out[8];
assign in[10] = out[9];
assign in[11] = out[10];
assign in[12] = out[11];
assign in[13] = out[12];
assign in[14] = out[13];
assign in[15] = out[14];
assign in[16] = out[15];
assign in[17] = out[16];
assign in[18] = out[17];
assign in[19] = out[18];
assign in[20] = out[19];
assign in[21] = out[20];
assign in[22] = out[21];
assign in[23] = out[22];
assign in[24] = out[23];
assign in[25] = out[24];
assign in[26] = out[25];
assign in[27] = out[26];
assign in[28] = out[27];
assign in[29] = out[28];
assign in[30] = out[29];
assign in[31] = out[30];
assign cout = out[31];
endmodule
//----------------------------------------------
// Arithmetic module. Adds if ctrl = 0, subtracts if ctrl = 1
module yArith(z, cout, a, b, ctrl);
output [31:0] z;
output cout;
input [31:0] a, b;
input ctrl;
wire [31:0] notB, tmp;
wire cin;
assign notB = ~b;
assign cin = ctrl;
yMux #(.SIZE(32)) mux(tmp, b, notB, ctrl);
yAdder adderSubtractor(z, cout, a, tmp, cin);
endmodule
//----------------------------------------------
TESTER:
module labL;
reg [31:0] a, b;
reg [31:0] expect;
reg [2:0] op;
wire ex;
wire[31:0] z;
reg ok, flag;
yAlu mine(z, ex, a, b, op);
initial
begin
repeat(10)
begin
a = $random;
b = $random;
op = 3'b101;
//flag = $value$plusargs("op=%d", op);
#10;
// ERROR CASE
if (op === 3'b011)
$display("Error!");
else if (op === 3'b111)
$display("Error!");
// ARITHM CASE
else if(op === 3'b010)
expect = a + b;
else if(op === 3'b110)
expect = a + ~b + 1;
// AND CASE
else if(op === 3'b000)
expect = a & b;
else if (op === 3'b100)
expect = a & b;
// OR CASE
else if (op === 3'b001)
expect = a | b;
else if (op === 3'b101)
expect = a | b;
// DONE WITH CASES;
#5;
if (expect === z)
$display("PASS: a=%b, b=%b, op=%b, z=%b", a, b, op, z, ex);
else
$display("FAIL: a=%b, b=%b, op=%b, z=%b, expect=%b", a, b, op, z, expect);
end
$finish;
end
endmodule
Your yMux4to1 does not drive the z output, so that's why you see 'zzz' as the output.
This means undriven/high-impedance.
You should be able to use a waveform viewer/simulator to trace your outputs (much better than using print statements).
You are getting a high-impedance signal out on Z. This means that your output Z is not driven. You should step through your design in simulation and put traces on your control signals and Z. Your IDE should support this. You most likely do not have the design wired up correctly so it's important to check your datapath and make sure all inputs/outputs are properly connected.