Say tp0 is the lowest hierarchy, all real code omitted just leave a parameter
module tp0;
parameter num=0;
endmodule
Where it is instanced
module tp
(
input id
);
//what I want to do is below code, but this will not pass compile since parameter need this is a constant!
tp0 #(id) tp_i();
endmodule
id is an input but this is actually be assign as fixed value in tp's instance.
module tb
.....
tp tp_0
(
.id(0)
);
tp tp_1
(
.id(1)
);
....
endmodule
How to resolve this if the pass of id is such a way and need to be passed into a parameterized module tp0?
Thanks
Verilog is a RTL language, which means that you are describing hardware that will be eventually synthesized.
When you use a parametrized instance, the parameter is analyzed at compile time and that affects how in synthesis the code will be transformed. The parameter NUM (capital letters suggested for params) usually refers to number of bits or instances, thus it needs to be known to calculate for example how many flip-flops should be instanced.
An input is something that dynamically changes while the application is running (ASIC or FPGA), thus it cannot be defined at compile time by connecting it to a parameter.
In your case, the best solution is to add a parameter also to tp module and then connect it to tp0.NUM, but it is not clear from this example what is your target or if this code is for simulation only (various quick alternatives) or if needs to be synthesized.
An elaborate alternative might be that in tp module you instance two submodules with each value of the parameter (0 and 1) and then mux the outputs depending on the value of id.
module tb();
tp tp0 (.id('0));
tp tp1 (.id('1));
endmodule
module tp(
input logic id
);
logic out0, out1, out_f;
tp_sub #(0) tp_0(.out(out0));
tp_sub #(1) tp_1(.out(out1));
always_comb begin
if (id) begin
out_f = out1;
//or other things
end
else begin
out_f = out0;
//or other things
end
end
endmodule
Related
I am trying to change a c++ code into verilog HDL.
I want to write a module that changes one of its inputs. (some how like call by reference in c++)
as I know there is no way to write a call by reference module in verilog (I can't use systemverilog)
Here is a code that I wrote and it works. are there any better ways to do this?!
my problme is that the register I want to be call by reference is a big array. this way duplicates the registers and has a lot of cost.
module testbench();
reg a;
wire b;
reg clk;
initial begin
a = 0;
clk = 0;
#10
clk = 1;
end
test test_instance(
.a(a),
.clk(clk),
.aOut(b)
);
always#(*)begin
a = b;
end
endmodule
module test(
input a,
input clk,
output reg aOut
);
always #(posedge clk) begin
if (a == 0)begin
a = 1;
aOut = a;
end
end
endmodule
Verilog is not a software programming language; it is a hardware description language. The inputs to a module are pieces of metal (wires, tracks, pins); the outputs from a module are pieces of metal. If you want a port that is both an input and an output you can use an inout. However, inout ports are best avoided; it is usually much better to use separate inputs and outputs.
A Verilog module is not a software function. Nothing is copied to the inputs; nothing is copied from the outputs. A Verilog module is a lump of hardware: it has inputs (pieces of metal carrying information in) and outputs (pieces of metal carrying information out).
Your are right to say that you can use either pass-by-copy or pass-by-reference in SystemVerilog. If you wish to pass a large data structure into a function or into/out of a task, then passing by reference may save simulation time.
By reference means by address, so to translate this to hdl directly you would either need to provide a way for the module to get on that bus and perform transactions based on that address.
Or better, if you need this as an input take each of the items in the struct and make individual inputs from them. If it is pass by reference because it is an output or is also an output, then you create individual outputs for each of the items in the struct. The module then distinguishes between the input version of that sub item and output version of that sub item.
my.thing.x = my.thing.x + 1;
becomes something like
my_thing_x_output = my_thing_x_input + 1;
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;
I would like to write a module with a variable number of inputs, i.e. depending on some parameter, the result would be:
module my_module #(LENGTH)(
input clk,
input rst_n,
input [LENGTH-1:0] data_1
);
//...
endmodule
or
module my_module #(LENGTH)(
input clk,
input rst_n,
input [LENGTH-1:0] data_1,
input [LENGTH-1:0] data_2,
input [LENGTH-1:0] data_3
);
//...
endmodule
Would it be possible to do this in Verilog or Systemverilog or would I have to write a script, let's say in Python, in order to generate the code for a specific module with fixed number of inputs? (it might be more than 1000 inputs)
There are no variable number of ports in SystemVerilog, but you could use a port that is a parameterized array.
module my_module #(int LENGTH, DEPTH)(
input clk,
input rst_n,
input [LENGTH-1:0] data[DEPTH]
);
//...
endmodule
Otherwise, you would need to use a script to generate the code.
Use a two dimensional input with a parameterized size. Added a generate for loop that can be used to set signals individually. Although many operations can be done with smart array operations.
module my_module #(SIZE, LENGTH)(
input clk,
input rst_n,
input [SIZE-1:0][LENGTH-1:0] data_in_array,
output [SIZE-1:0][LENGTH-1:0] data_out_array
);
genvar N;
generate for (N=0; N<SIZE; N++) begin :la_coolOps
//Do cool operations here. For example instantiate a module for every data_in
end
//...
endmodule
Edit:
As Mehran Torki points out: The syntax above will work for SystemVerilog only. Verilog does not allow for multiple packed arrays. Use input [LENGTH*SIZE-1:0] data_in_array.
I would add to these other answers that ports are just groupings of wires. While having 3, 1-bit wires named a, b, and c might be easier to read and understand, there is no physical/logical difference between a single, 3-bit wire abc, where abc[0] corresponds to a, abc[1] corresponds to b, and abc[2] corresponds to c.
So, you can always just expand or shrink a single (or multiple) signal(s) to get however many bits you need. It may not be as neat, but it will work. In the receiving module, you can then part-select the bus in whatever manner you like. So, you could have one really long wire the shrinks or expands (wire [(SOME_PARAM*8)-1:0] my_input_wire), or with SystemVerilog an array (wire [7:0] my_input_wire[0:SOME_PARAM-1])
If this is just testbench/verification code, the other thing you could do in SystemVerilog is use a dynamic array
As others said, there is no direct way to do this, but another workaround is to use SystemVerilog interfaces, where you define all the inputs that you want in the interface definition and inside the module only use the ones that correspond to the parameter. Below is a sample:
module my_module #(LENGTH)(
input clk;
input rst_n;
output o;
interface i_data;
);
logic outValue;
generate
case (LENGTH) //Based on the value of LENGTH, use corresponding data
1: outValue = i_data.data_1;
2: outValue = i_data.data_1 + i_data.data_2;
3: outValue = i_data.data_1 + i_data.data_2 + i_data.data_3;
endcase
endgenerate
always #(posedge clk) begin
if (~rst_n)
o <= '0;
else
begin
o <= outValue;
end
endmodule
You can still use a parameterized array for data and a for-generate loop if your outputs are similar.
With System verilog we can import a package instead of having parameterization and define types in the package to be used in the portlist.
module mymodule
import mymodule_pkg::*;
(
input portlist_t portlist
);
endmodule
And define multiple copies of the package with different variants of the port list and compile whichever version is required. eg
package mymodule_pkg;
localparam LENGTH=5;
typedef struct packed {
logic [LENGTH-1:0] data_1,
logic [LENGTH-1:0] data_2,
logic [LENGTH-1:0] data_3
} portlist_t;
endpackage
As with the interface solution there will be situations where you run into issues, like having different iterations of the module instantiated together.
I have a Lattice Diamond project for an SPI multiplexer, which has the following module definition:
module spimux
(
input bmck,
input bssel,
input bmosi,
output bmiso,
input[3:0] a,
output[13:0] mck,
output[13:0] ssel,
output[13:0] mosi,
input[13:0] miso,
output reg[7:0] LED
);
OutputMux bmiso_mux (
.clk(osc_clk),
.out(bmiso),
.a(a),
.in(miso)
);
// the idea here is that on each rising clock edge, the module will take
// the 4-bit address a and then set *one* of the 14 bits in "in". One
// problem I see is that I don't prevent an invalid address of 0b1111 or
// 0b1110 from getting used.
module OutputMux
(
input clk,
output reg out,
input[3:0] a,
input[13:0] in
);
reg mask;
always #(posedge clk) begin
// I tried this and it didn't help my situation
//out <= (in & (14'b1 << a));
// so I tried to assign to a temp variable and then do the bitmasking.. no change.
mask = 14'b1 << a;
out <= (in[13:0] & mask);
end
endmodule
endmodule
When I go into the Spreadsheet View to assign my pins, not all of them show up in the Signal Name droplist. For example, it looks like this:
You'll see that miso[0] is in there as an Input Port, but all of the other 13 miso bits are not. In addition, bmck, bssel, and bmosi are missing. They have not yet been assigned to any other pins, so can anyone explain why they would not be there?
Thanks to Qiu for getting me going in the right general direction. I should have guessed that the signal name list is generated after compiling the Verilog code, so if the output/input isn't getting used, you won't need to map it to a pin.
Using compileonline.com, I was able to quickly iterate over my Verilog logic statements and figure out where the problem came from. For miso, I was able to make them appear by changing my always block to look like this:
always #(posedge clk) begin
out = (in[13:0] & (14'b1 << a)) > 0;
end
The idea here is really simple -- out of all of the MISO inputs entering the FPGA, we only want to look at the one coming from the SPI device that is currently selected (identified by address a). We just need to set out to the value of the bit identified by a. After masking, the value is going to be 0 or !0, so we just write this to out.
I wanted to use a reduction operator, but the online compiler didn't seem to work with this notation, so I just compared to 0 instead, which seems to work. I still have to test this on hardware.
I am trying to write an "inverter" function for a 2's compliment adder. My instructor wants me to use if/else statements in order to implement it. The module is supposed to take an 8 bit number and flip the bits (so zero to ones/ones to zeros). I wrote this module:
module inverter(b, bnot);
input [7:0] b;
output [7:0]bnot;
if (b[0] == 0) begin
assign bnot[0] = 1;
end else begin
assign bnot[0] = 0;
end
//repeat for bits 1-7
When I try and compile and compile it using this command I got the following errors:
vcs +v2k inverter.v
Error-[V2005S] Verilog 2005 IEEE 1364-2005 syntax used.
inverter.v, 16
Please compile with -sverilog or -v2005 to support this construct: generate
blocks without generate/endgenerate keywords.
So I added the -v2005 argument and then I get this error:
vcs +v2k -v2005 inverter.v
Elaboration time unknown or bad value encountered for generate if-statement
condition expression.
Please make sure it is elaboration time constant.
Someone mind explaining to me what I am doing wrong? Very new to all of this, and very confused :). Thanks!
assign statements like this declare combinatorial hardware which drive the assigned wire. Since you have put if/else around it it looks like you are generating hardware on the fly as required, which you can not do. Generate statements are away of paramertising code with variable instance based on constant parameters which is why in this situation you get that quite confusing error.
Two solutions:
Use a ternary operator to select the value.
assign bnot[0] = b[0] ? 1'b0 : 1'b1;
Which is the same as assign bnot[0] = ~b[0].
Or use a combinatorial always block, output must be declared as reg.
module inverter(
input [7:0] b,
output reg [7:0] bnot
);
always #* begin
if (b[0] == 0) begin
bnot[0] = 1;
end else begin
bnot[0] = 0;
end
end
Note in the above example the output is declared as reg not wire, we wrap code with an always #* and we do not use assign keyword.
Verliog reg vs wire is a simulator optimisation and you just need to use the correct one, further answers which elaborate on this are Verilog Input Output types, SystemVerilog datatypes.