I'm very new to verilog and i'm just starting to understand how it works.
I want to manipulate an input to a module mant[22:0], in an always block but I am not sure how to go about it.
module normalize(mant,exp,mant_norm,exp_norm);
input [22:0]mant;
input [7:0]exp;
output [22:0]mant_norm;
output [7:0]exp_norm;
reg mantreg[22:0];
reg count=0;
always#(mant or exp)
begin
mantreg<=mant; //this gives an error
if(mant[22]==0)
begin
mant<={mant[21:0],1'b0};//this also gives an error
count<=count+1;
end
end
endmodule
so i have to shift the mant register if the bit22 is zero and count the number of shifts. I am so confused about when to use reg and when to use wire and how to do the manipulation. please help let me know how to go about it.
As you can see in your code you are assigning vector value (mant) to array of 23(mantreg). Instead you should declare mantreg as reg [22:0] mantreg (which is vector of 23 bit).
Wire type variable can not be assigned procedurally. They are used only in continues assignment. Other way around reg varible can only be procedural assigned.
For that try to read out LRM of Verilog .
Related
Image. I am trying to solve a problem that was written below. I confused why my output was in undefined state.
A "population count" circuit counts the number of '1's in an input vector. Build a population count circuit for a 255-bit input vector.
module top_module(
input [254:0] in,
output [7:0] out );
reg [7:0] counter=8'b0;
reg [7:0] counter_next=8'b0;
always # (*)
begin
counter=counter_next;
end
always # (*)
begin
for (int i=0; i<$bits(in);i++)
counter_next=counter+in[i];
end
assign out=counter;
endmodule
confused why my output was in undefined state.
All HDL variables are undefined ('X' or 'U') until you give them a value.
Then, you add a value to that undefined value counter_next=counter+in[i]; which still gives undefined. So it stays that way.
Also you are using 'counter=counter_next' which suggest you have seen some existing HDL code and you are trying to copy it, but you do not understand why it is implemented that way. The 'next' system is used when there is a clock. You do not have a clock and as such it is superfluous here.
The code you are looking for is probably something like this:
output reg [7:0] out
always #( * )
begin
out = 0;
for (int i=0; i<$bits(in);i++)
out =out+in[i];
end
Note that I am not using an extra variable counter here. All I do is make 'out' a reg type so I can use it directly.
I am interested to know what will these lines synthesize to? I am designing a direct mapped cache and using assign to separate my index, offset and tag bits. Will it synthesize to a demultiplexer? I did not mention the whole code since I just want to know how the assign statements will look after synthesis. In 8085 programming, I had read the term as 'address demultiplexing" so it was confusing.
module cache
( input bit clk,
input bit rst,
input logic [15:0] address,
input logic valid_in,
input logic compare,
input logic wr,
input logic enable,
input logic write_through,
output logic dirty,
output logic [4:0] tag_out,
output logic [15:0] data_out,
output logic valid_out,
output hit
);
logic [7:0] index;
logic [1:0] offset;
logic [4:0] tag_in;
assign offset = address[1:0];
assign index = address[9:2];
assign tag_in = address[15:10];
endmodule
The above code will just simply get synthesized as wire's, since there are only assignments.
I am not sure what de-multiplexing logic you are trying to create, but generally for a de-multiplexer you need to have a select signal based on which you decode which output should be enabled.
An example for a 1:2 de-multiplexer logic is given below
module demux_1_2(
input [3:0] Q,
input Sel,
output reg [3:0] D1,
output reg [3:0] D2
);
always#(*)
begin
if(~Sel) begin
D1 = Q;
D2 = 0;
end else begin
D1 = 0;
D2 = Q;
end
end
endmodule
Since there are no Boolean or arithmetic operators on the RHS of the assign, these statements just become conveniently named references for part selects of the address input. This is the same thing that happens when you instantiate a module and connect to its ports - signals can go through a name change. In fact, you could have you could have written your address input port declaration as
input .address({tag_in,index,offset}),
You still connect the address port when instantiating this module, but inside the module, it only has tag_in, index, and offset available to reference, not address.
SystemVerilog has the alias construct to make it more obvious that you are just creating a convenient names for a signal, instead of declaring another set of signals and using the assign statement.
alias offset = address[1:0];
alias index = address[9:2];
alias tag_in = address[15:10];
I have made two verilog modules. The first one takes a nine-bit number and returns the position of first occurrence of 1 in it.
module findPositionOf_1(
input [8:0] data,
output reg [3:0] position
);
always #(data)
begin
if(data==9'b0000_00000)
position=4'b0000;
else if(data[0]==1)
position=4'b0000;
else if(data[1]==1)
position=4'b0001;
else if(data[2]==1)
position=4'b0010;
else if(data[3]==1)
position=4'b0011;
else if(data[4]==1)
position=4'b0100;
else if(data[5]==1)
position=4'b0101;
else if(data[6]==1)
position=4'b0110;
else if(data[7]==1)
position=4'b0111;
else if(data[8]==1)
position=4'b1000;
end
endmodule
The second module is returning the second occurrence of 1. It is calling the first module first changing that bit to zero and again finding the occurrence of 1.
module findPositionOf_2nd_1(
input [8:0] r1_data,
output [3:0] position1
);
reg [3:0] pos,pos2;
reg [8:0] temp;
integer i;
always #(r1_data)
begin
findPositionOf_1 f1(.data(r1_data), .position(pos));
i=pos;
temp=r1_data;
temp[i]=0;
findPositionOf_1 f2(temp,pos2);
if(pos2==4'b0000)
position1=0;
else
position1=pos2;
end
endmodule
I am getting the following errors during compilation. Please help.
Checker 'findPositionOf_1' not found. Instantiation 'f1' must be of a
visible checker.
A begin/end block was found with an empty body. This
is permitted in SystemVerilog, but not permitted in Verilog. Please
look for any stray semicolons.
By the way you write code it seems like you've not completely grasped how verilog(and other HDL languages) is different from "normal", procedural, coding.
You seem to assume that everything inside your always# block will execute from top to bottom, and that modules are similar to functions. This is not the case. You need to think about how you expect the hardware to look when you've designed your module.
In this case you know that you want two findPositionOf_1 modules. You know that you want the result from the first (u_f1) to affect the input of the second (u_f2). To do this, instantiate the two modules and then determine the interconnect between them.
We can create a vector with a 1 in position pos by left-shifting '1 pos number of times (1<<pos). By xor-ing the bits together, the statement r1_data ^ 1<<pos will remove the unwanted 1.
module findPositionOf_2nd_1(input [8:0] r1_data, output [3:0] position1 );
wire [3:0] pos,pos2;
wire [8:0] temp;
findPositionOf_1 u_f1(.data(r1_data), .position(pos));
findPositionOf_1 u_f2(.data(temp), .position(pos2));
assign temp = r1_data ^ (1<<pos);
assign position1 = pos2;
endmodule
You have instantiated your module inside an always block which is a procedural block, which is syntactically incorrect. Secondly, you have used your first module as a function call, which is not permitted. As said, you need to have a separate testbench, where you can connect your both modules and check. Make the position of occurance of 1st one as input to the findPositionOf_2nd_1 module. For your question, perhaps this should help
Why can't I instantiate inside the procedural block in Verilog
module tff(t,i,qbprev,q,qb);
input t,i,qbprev;
output q,qb;
wire q,qb,w1;
begin
assign w1=qbprev;
if(w1==1)begin
not n1(i,i);
end
assign q=i;
not n2(qb,i);
end
endmodule
module counter(a,b,c,cin,x0,x1,x2);
input a,b,c,cin;
output x0,x1,x2;
reg a,b,c,x0,x1,x2,temp,q,qb;
always#(posedge cin)
begin
tff t1(.t(1) ,.i(a),.qbprev(1),.q(),.qb());
x0=q;
temp=qb;
tff t2(.t(1) ,.i(b),.qbprev(temp),.q(),.qb());
x1=q;
temp=qb;
tff t3(.t(1) ,.i(c),.qbprev(temp),.q(),.qb());
x2=q;
a=x0;
b=x1;
c=x2;
end
endmodule
This is my code in verilog. My inputs are - the initial state - a,b,c and cin
I get many errors with the first of them being "w1 is not a constant" What doesn this mean?
I also get error "Non-net port a cannot be of mode input" But I want a to be an input!
Thank you.
Modules are instantiated as pieces of hardware. They are not software calls, and you can not create and destroy hardware on the fly therefore:
if(w1==1)begin
not n1(i,i);
end
With that in mind I hope that you can see that unless w1 is a constant parameter, and this is a 'generate if' What your describing does not make sense.
instance n1 is not called or created as required, it must always exist.
Also you have the input and output connected to i. i represent a physical wire it can not be i and not i. these need to be different names to represent different physical wires.
In your second module you have :
input a,b,c,cin;
// ...
reg a,b,c; //...
Inputs can not be regs as the warning says, just do not declare them as regs for this.
input a,b,c,cin;
output x0,x1,x2;
reg x0,x1,x2,temp,q,qb;
Here is the declaration of the reg assignment
reg [5:0]R = {bi7 ,[15:11]RGB}; //bi7 is a parameter
but at the last line of the module i get this error where it points at the same reg assignment.
ERROR:HDLCompiler:69 - "path.v" Line 58: <R> is not declared.
Can anyone help me with this , cause my whole experience with verilog is just a book :(
In verilog, you can only assign a value to a reg in always or initial blocks. You've also got the bit range for stripping bits from you RGB bus on the wrong side of the bus name.
reg [5:0] r;
always #(RGB) begin
r = {bi7, RGB[15:11]};
end
Note that in verilog, parameter names such as bi7 in your code, are usually defined and written in UPPER CASE to make them easy to pick out.