Shift Registers Verilog

Question

I am very new to HDL language. I have a question about how to program a shift register. (i know i shift to the other direction). Why does the book use wire[N-1:0] r_next? what's drawback of my implementation? thanks

my first try is as following

module lesson04#(parameter N=8)(
    input wire clk, reset,
    input wire data,
    output wire out
);

reg [N-1: 0] r_reg;


always @(posedge clk or negedge reset)
begin
    if(!reset)
         r_reg =0;
    else 
        r_reg[0]=data;
        r_reg = r_reg<<1;
end

assign out =r_reg[N-1];
endmodule 

but the book gives:

module lesson04#(parameter N=8)(
    input wire clk, reset,
    input wire data,
    output wire out
);

reg [N-1: 0] r_reg;
wire[N-1:0] r_next;

always @(posedge clk or negedge reset)
begin
    if(!reset)
         r_reg =0;
    else 
        r_reg <= r_next;
end

assign r_next= {data, r_reg[N-1:1]};
assign out =r_reg[N-1];
endmodule

Answer 1

First of all, don't forget your begin-ends around sections of code:

else begin
     r_reg[0]=data;
     r_reg = r_reg<<1;
end

Without this, only r_reg[0]=data will be in the else clause of the if statement. This will work, but is considered bad style due to the blocking statements in a sequential logic description...

Second, for modeling sequential blocks, use nonblocking assignments (<=) or your calculations may 'fall through' (google nonblocking vs blocking for more info). Your example may very well work (did you try it in a simulator?) but if things get more complicated and more variables are added things can break.

always @(posedge clk or negedge reset)
begin
    if(!reset)
         r_reg <= 0;
    else begin // This is horrible! Don't write code like this!
        r_reg[0] = data;     // blocking
        r_reg <= r_reg<<1;   // non-blocking
    end
end

For the above reason, it is sometimes recommended that combo logic is separated from sequential logic so that you can write nonblocking assignments to registers in sequential blocks, and blocking in combo blocks and never have to worry about the scheduling.

To code in this way, you need to calculate what the next output should be using the current state, hence the r_next bus in the answer. I think it tends to help the synthesis tool out too if all the flip-flops are separated from surrounding combo logic in this way.

Also, if your reset is active low (ie LOW resets ) it should be named as such, eg resetb or reset_n.

Answer 2

Your implementation produces quite a different output from the book's. You should prove this to yourself by constructing a simple testbench to drive your inputs and run a simulation. You will see that the book's output shifts the input data by a single clock cycle, whereas your output shifts the input data by eight clock cycles.

By the way you have indented your always block, I am led to believe that it is not what you wanted. This is how your block really behaves:

always @(posedge clk or negedge reset)
begin
    if(!reset) begin
         r_reg =0;
    end else begin
        r_reg[0]=data;
    end
    r_reg = r_reg<<1;
end

I always explicitly use the begin/end keywords in if/else statements to avoid this confusion.

The way it simulates, r_reg is always 0 because you clobber the 1st assignment (r_reg[0]=data;) with the 2nd (r_reg = r_reg<<1;). Another difference is that the book assigns data to the MSB of the shift register, but you assign it to the LSB.

If you are using decent linting and synthesis tools, you would probably get a bunch of warnings for your code. This would alert you to make some changes.